Class 5 General Knowledge - Basic Science

Why do the glaciers move?

While pressing in hands a bit of snow melts and when pressure is released again freezing point reaches and the melted water freezes building the particles of snow.

Atom of which element has two protons and two electrons ?

If an atom has

$2$ nuculear protons, it means the atom is helium.

The identity of an atom is determined by its atomic number,

$Z$ , which is equal to the number of positively charged particles called protons, present inside the nucleus of the atom. If

$Z=1$ , the atom is hydrogen. If

$Z=2$ , the atom is helium. If

$Z=3$ , the atom is lithium, and so on.

The number of protons in an element is always equal to the number of electrons present in it.

Therefore, the atom of the element that has two protons and two electrons is helium.

How do you write an equation for the reaction between zinc metal and hydrochloric acid (aqueous) producing zinc chloride (aqueous) and hydrogen gas?

The required reaction willbe:

$Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$

A man descending by parachute does it slowly. Why?

It is due to variation in pressure of the atmosphere.

A tennis ball is thrown up and reaches a height of $4.05 m$ before coming down. What was its initial velocity? How much total time will it take to come down? Assume $g = 10\ m/s^{2}$ .

$H_{max} = \dfrac {u^{2}\sin^{2} 90^{\circ}}{2g}$

$4.05 = \dfrac {u^{2}\times 1}{2\times 10}$

$u^{2} = \dfrac {4.05}{20} = 0.2025$

$u = \sqrt {0.025} = 0.45\ m/s$

$T = \dfrac {24\sin 90^{\circ}}{g} = \dfrac {2\times 0.45}{10}$

$= 0.09\sec$ .
1877861_1900163_ans_023b0493db224177a1857ce63ee0171f.png

Consider the situation shown in figure. The is released from rest and the block of mass $1.0\ kg$ is found to have a speed $0.3\ m/s$ after it has descended through a distance of $1\ m$ . Find the coefficient of kinetic friction between the block and the table.

$T-f_{k}=4\times (2a)...(i)$

$10-2T=1\times a....(ii)$
from eqn (i) & (ii)

$27-2f_{k}+10-27=16a+a$

$-2f_{k} +10=17\times a$

$-u_{k}mg+10=17\times \dfrac {0.09}{2}$

$-u_{k}(40)=0.765-10$

$-u_{k} =-\dfrac {-9.235}{40}$

$u_{k} =0.23$
the coefficient of kinetic friction between the block and the table is

$u_{k}=0.23$

A cyclist speeding at $18\ km/h$ on a level road takes a sharp circular turn of radius $3\ m$ without reducing the speed. The co-efficient of static friction between the tyres and the road is $0.1$ . Will the cyclist slip while taking the turn?

$v=18\ km/h=18\times \dfrac{5}{18}=5\ m/sec$

$R=3\ m$

$f=\mu mg$

$=0.1\times m\times 10$

$f=m$ ....(1)
for no slipping

$f\ge \dfrac{mv^{2}}{R}$

$m > \dfrac{mv^{2}}{R}$

$\dfrac{v^{2}}{R} < 1\Rightarrow \dfrac{(5)^{2}}{3}=\dfrac{25}{3}$
So the cylist will slip
1978294_1900193_ans_40877e4bcbaa458985ce5a0ed68f69c5.png

Prove that the angular momentum of a satellite of mass $M_{s}$ revolving round the earth of mass $M_{e}$ in an orbit of radius $r$ is equal to $[GM_{e}M_{s}^{2}r]^{1/2}$ .

$F=\dfrac{GM_{e}M_{s}}{r^{2}}=\dfrac{MS_{2}}{r}$
(

$G$ . force)
(centripital force)

$=\dfrac{GM_{e}}{r}=v^{2}$

$v=\sqrt{\dfrac{GM_{e}}{r}}$
Angular momentum

$=M_{S}\cup r$

$=MS\sqrt{\dfrac{GM_{e}}{r}}\times r$

$=\sqrt{GM_{e}M_{s}^{2}r}$

In the $x-t$ graph below, indicate the time when the average velocity is equal to the instance velocity.

nstantaneous

$\Rightarrow \dfrac{dx}{dt}=\dfrac{x_1-X_6}{t_1-0}$

$avg=\dfrac{displacement }{time}$

$=\dfrac{X_1-X_0}{t_1-0}$

A car of mass $m$ is coasting on a horizontal surface. There is no friction between the car and the surface, but a force of air resistance acts on the car of magnitude $F_{drag}(v) =kv^2$ . The initial velocity of the car is $v_0$ .

Let at any instant reativity of car is

$v$

$\displaystyle \int =kv^2$

$ma =kv^2$

$m \dfrac {dv}{dt}=kv^2$

$\displaystyle \int_{v_0}^{v\ dv } = \displaystyle \int_{t=0}^{t }\dfrac km v^2\ dt$

$\displaystyle \int_{v_0}^v \dfrac {dv}{dt} = \dfrac km \displaystyle \int_0^t dt$

$\left[-\dfrac {1}{v}\right]_{v_0}^v =\dfrac {k}{m}t$

$-\left(\dfrac {1}{v}-\dfrac {1}{v_0}\right)=\dfrac {k}{m}t$

$\dfrac {1}{v}-\dfrac {1}{v_0}=-\dfrac {k}{m}t$

$\dfrac {1}{v}=\dfrac {1}{v_0}-\dfrac {k}{m}t =\dfrac {m-v_0 kt}{v_0 m}$

$v=\dfrac {mv_0}{m-v_0 kt}=\left[\dfrac {v_0}{1-\dfrac {v_0}{m}kt}\right]$

As shown in the figure a person is pulling a mass 'm ' form ground on a fixed rough hemispherical surface upto the top of the hemisphere with with the help of a light inextensible string find the work done by tension in the string if radius of hemisphere is R and friction co-efficient is $\mu$ Assume that the block is pulled with negligible velocity

By energy conservation we have

$W-R = mg R+ \mu mg [ - cos \theta]^{90}_0$

$= mg R - \nu mg [ cos 90 - cos 0 ]$

$= mgR ( 1+\mu)$

Therefore this the required work done.

A cylinder and a wedge of same masses with vertical face , touching each other move along two smooth inclined planes forming the some angle and $\beta$ respectively with the horizontal determine the force of normal N (in newton) exerted by the wedge on the cylinder , neglecting the friction between them if $m = \frac {1}{ \sqrt {3} } kg \alpha = 60^0 , \beta = 30^0$ and $g = 10 m/s^2$

$N_1 cos \beta = mg$

$N_ 1 sim \beta = N$

$N_1 cos 30^0 = \frac {1}{ \sqrt {3} } \times 10$

$\Rightarrow N_1 = \frac {20}{3}$

$N_1 sin 30^0 = N$

$\Rightarrow N = \frac {20}{3} \times \frac {1}{2} = \frac {10}{3} N$
1978307_1900572_ans_e74e7ae80529477ea25f8497c143bd4f.png

Diagram shows a velocity-times graph for a car starting from rest. The graph has three sections $AB,BC$ and $CD$ .
In which section, car has a zero acceleration.

here

$\theta_1 < \theta_2$ hence

$a= \tan \theta$

$A\ ar\ \theta < \tan \theta_2$ hence

$a_1 < a_2$
acceleration

$<$ Retardation
in

$BC; v=constant$
hence

$a=0$
Area of

$v-T$ represented the distance

A car with a gun mounted on it is kept on horizontal friction less surface total mass of car , gun and shell is 0 kg mass of each shell is 1 kg if shell is fired horizontally with relative velocity 100 m/ sec with respect to gun what is the recoil speed of car after second shot in nearest integer ?

Center of man remain const

$o = 49 mv + m \times 200$

$\Rightarrow V = \frac { -2 \omega}{49} m/ s$ towards left

another fired

$C_1 = \frac {200}{48} mis$ towards left

Now velocity of car wrt earth

$\Rightarrow \frac { 2 \omega}{49} + \frac { 2 \omega}{48}$

$= 8.24 m/s$

Nearest integer is 8.

A cylinder and a wedge of same masses with vertical face , touching each other move along two smooth inclined planes forming the some angle and $\beta$ respectively with the horizontal determine the force of normal N (in newton) exerted by the wedge on the cylinder , neglecting the friction between them if $m = \frac {1}{ \sqrt {3} } kg \alpha = 60^0 , \beta = 30^0$ and $g = 10 m/s^2$

$N_1 cos \beta = mg$

$N_ 1 sim \beta = N$

$N_1 cos 30^0 = \frac {1}{ \sqrt {3} } \times 10$

$\Rightarrow N_1 = \frac {20}{3}$

$N_1 sin 30^0 = N$

$\Rightarrow N = \frac {20}{3} \times \frac {1}{2} = \frac {10}{3} N$

1874638_1900571_ans_96d555e369804df5b22a53204ce07caa.png

A particle if given a certain velocity v at point P as shown on a hemispherical smooth surface find the value of v (in m/s) such that when particle reaches Q , the normal reaction of surface becomes equal to particle's weight $[ R = 1.6 m , g = 10 m/s^2 ]$

From figure

At top

$N - mg - \frac { -mv^2}{R}$

$N = mg \frac {mv^2}{R} = 0 \Rightarrow V = 0$

is the particle steps at Q

Covering energy

$\frac {1}{2} mv^2 = mg ( R - R co 60^0 )$

$\frac {1}{2} mv^2 = \frac {mg R}{2} \Rightarrow V = \sqrt {Rg}$

$V = \sqrt { 1.6 \times 10}$

$= \sqrt {16}$

$= 4 m/s$

Diagram shows a velocity-times graph for a car starting from rest. The graph has three sections $AB,BC$ and $CD$ .
Is the magnitude of acceleration higher or lower than that of retardation? Give Reason.

here

$\theta_1 < \theta_2$ hence

$a= \tan \theta$

$A\ ar\ \theta < \tan \theta_2$ hence

$a_1 < a_2$
acceleration

$<$ Retardation
in

$BC; v=constant$
hence

$a=0$
here are aof

$AB \angle$ area of

$BC$

$(S_1=)\dfrac{1}{2} vx + \angle v \times t (=S_2)$

$\dfrac{S_1}{S_2} \dfrac{S_2}{S_1} =\dfrac{vA}{\dfrac{1}{2} VA}=\dfrac{2}{1}$

The position of a particle along the $x-$ axis is given in centimeter by $x=9.75+1.50 t^3$ , where $t$ is in second. Consider the time interval $t=2\ s$ to $t=3\ s$ and calculate the average velocity.

$x=9.75+1.5t^3$

Average velocity

$=\dfrac {displacement}{times}$

$\therefore$ Displaceemnt between

$t=2$ to

$t=3$

$(9.75+1.5t^3)_3-(9.75+1.5t63)_2$

$=1.5(3)^3 -(1.5)(2)^2$

$=1.5\left\{27-8\right\}=28.5$

=Average velocity

$=\dfrac {25.5}{3-2}=28.5\ m/s$

A person moves through a distance of $20\ m$ due to east, then moves through a distance $20\ m$ due north, then turns west and covers a distance of $10\ m$ and then turns south and covers a distance of $10\ m$ . Find the displacement.

According to figure we have

Using PGTS

$H^2 =10^2+10^2$

$H=\sqrt {100+100}$

$H=\sqrt {200}$

$=10\sqrt 2 m$

$H=\sqrt {20^2 +20^2}$

$=\sqrt {400+400}$

$=\sqrt {2\times 400}$

$=20\sqrt {2}m$

1874831_1900646_ans_4153df5147304691a751779e8ce3afc4.png

Find moment of inertia of a uniform annular disc of mass 100 g, having inner radius 10 cm and outer radius 20 cm, about an axis passing through its centre and perpendicular to its plane.

We know that

$I = mK^2$

$m =$ mass of the disc

$k=$ radius of gyration

$I_{net} = m \times \dfrac{\left( R_{outer}^2 - R_{inner}^2\right)}{2}=0.1 \times \dfrac{\left( 0.2^2 - 0.1^2\right)}{2}= 2.5\times10^{-3}$

A boat with initial speed $v_0$ is showed by a frictional force $F=b\,exp (-\alpha v)$ .find the time after which it stops.

According to initial data we have

acceleration

$=\dfrac {F}{m}=\dfrac {b}{m}e^{-\alpha v}\ F=be^{-\alpha v}$

Now,

$\dfrac {dv}{dt}=a.c =\dfrac {F}{m}=\dfrac {b}{m}e^{-alpha v}$

$\displaystyle \int_0^v dv \times e^{\alpha v}=\displaystyle \int_0^t \dfrac bm dt$

$\dfrac {e^{\alpha v}}{\alpha} =\dfrac bm \times t +c$ . At

$t=0, v=0$

$\dfrac {e^{\alpha v_0}}{\alpha}=0+c \Rightarrow C: \dfrac {e^{\alpha v_0}}{\alpha}$

$\therefore (t)=\dfrac {b}{m}\times t+\dfrac {e^{\alpha v_0}}{\alpha}$ And

$x(t)=\displaystyle \int \dfrac {dv}{dt}\times (t)$

$x=\displaystyle \dfrac {ct}{m}+\dfrac {e^{\alpha v_0}}{\alpha}dt$

$(t)=\dfrac {bk}{2m}+\dfrac {e^{\alpha v_0}}{\alpha}t$

$t=0,$ at

$-\dfrac {e^{\alpha v_0}}{\alpha}\times \dfrac {m}{b}=t$

$\therefore t$ dill towards

$=\dfrac {5}{2m}\times \dfrac {e^{2\alpha v_0}}{\alpha^2}\times \dfrac {m^2}{b^2}+\left(\dfrac {e^{\alpha v^0}}{\alpha }\right)\times \left(\dfrac {-e^{\alpha v^0}}{\alpha }\times \dfrac mb\right)$

$=\dfrac {me^{\alpha v_0}}{2b \alpha^2}$

Two masses of $1\ kg$ and $5\ kg$ are attached to the end of a massless string passing over a pulleys of negligible weight. The pulley itself is attached to a light spring balance as shown in figure. The masses start moving during the interval, the reading of spring balance will be

On balancing the forces we get

$50-T=5a$ .....(i)

$T-10=11a$ ....(ii)

$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$

$50+10=6a$

$\therefore a=\dfrac{40}{6}=\dfrac{10}{3}ms^{-2}$

$\therefore T=a+10=\dfrac{20}{3}+10=10\ N$

$=\dfrac{50}{3}$

$\therefore$ Reading

$=29=\dfrac{100}{3}$

1874946_1900665_ans_7b89439073864236992c803b646291c3.png

The position of a particle along the $x-$ axis is given in centimeter by $x=9.75+1.50 t^3$ , where $t$ is in second. Consider the time interval $t=2\ s$ to $t=3\ s$ and calculate the instantaneous velocity when the particle is mid way between its position at $=2\ s$ and $t=3\ s$ .

Now Distance lowered between mid way of

$t=2,$ to

$t=3$

$=\dfrac {9.75+(1.5)(3)^3 + 9.75+(1.5)(2)^2}{2}$

$=\dfrac {x_1 +x_2}{2}$

$=9.75+\dfrac {1.5}{2}\left\{27+18\right\}$

$9.75+26.25=36$

thus at

$x=36$

$=36=9.75+1.57^3$

$\dfrac {26.25}{1.5}=t^3$

$t^3 =17.5$

$v=1.5\times 3 (t)^2$

$=1.5\times 3(2.6)^2$

$=30.42\ m/s$

The position of a particle along the $x-$ axis is given in centimeter by $x=9.75+1.50 t^3$ , where $t$ is in second. Consider the time interval $t=2\ s$ to $t=3\ s$ and calculate
Instantaneous velocity at $t=2\ s$ .

Instantaneous

$v=\dfrac {dx}{dt}$

$v=\dfrac {d}{dt}(9.75+1.57^3)$

On differentiating we get

$=1.5(3)t^2$

$=4.5t^2|_{t=2}$

$=4.5(2)^2 =18m/s$

Required velocity is 18m/s

A traffic police officer observes a fast moving car due to over speed officer starts his bike , accelerates uniformly to 90 kph in 8 s , and maintaining a constant velocity of 90 kph , overtakes the car 42 s after the car passed him if he overtakes the car after 18 s from the instant he starts , find the distance the officer travelled before overtaking the speed of the car

Let speed of car = V
time taken

$= 42 sec$

$V \times 42 = 350$

$V = \frac {350}{42}$

$V = \frac {25}{3} \times \frac {18}{5}$

$= 30 km / h$

Two thin uniform rod $A ( M, L)$ and $B ( 3M, 3L)$ are joined as shown find the MI about an axis passing through the centre of mass of system, of rods and perpendicular to the length

Mi of A about required axis

$= \frac {mL^2}{12} + , [ \frac {81}{2}]^2$

$= \frac {mL^2}{12} + \frac {gml^2}{4}$

$\frac {28 ml^2}{12}$

$= \frac {7}{3} ml^2$
now

$\frac { 3m(sL)^2}{ 12} + 3m [ L^2 ]$

$= \frac { 9 mc^2}{4} + 3ml^32 = \frac { 21 ml^2}{4}$

$= \frac {7}{3} + \frac {21}{4} ml^2$
1872555_1900846_ans_8032fd8aae2e48fcb417720d32620015.png

A 6.5 m long ladder rests against a vertical wall reaching a height of .0 m A 0 kg man stands half way up the ladder (a) find the torque of the force exerted by the man on the ladder about the upper end of the ladder (b) assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth find the force exerted by the ground on the ladder.

$t = 600 \times \frac { 6.5}{ 2} sin \theta$

$= 600 \times 6.52 \times \sqrt { 1 - cos^2 \theta }$

$\tau = 600 \times \frac { 6.5}{2} \times ( \sqrt { 1 - ( \frac {6}{6.5})^2 } ) = 735$

$\tau = 70 N - m$

$R_2 = mg = 60 \times 9.8 = 590$

$R_1 = mR_2$

$6.5 R_1 cos \theta = 6.09 sin \theta \times \frac { 6.5}{ 2}$

$R_1 = \frac {1}{26} tan \theta = \frac {1}{260} \times \frac { 2.6}{6}$

$R_1 = \frac {12.59}{6}$

$R_1 = 122.5 N$
1872633_1900913_ans_48837bb18aa94abcade07ad2568c78a8.png

If the particle is moving along the x-axis and the relation between acceleration and position (x) is given by a $= -9x\ m/s^{2}$ . If velocity of the particle at x = 3m is zero, then the velocity of particle at origin is:

Given

$a = -9x$

$\dfrac{dv}{dt} = -9x$

$\dfrac{dv}{dt} \times \dfrac{dx}{dt} = -9x$

$v\dfrac{dv}{dx} = -9x$

$\int^{0}_{v} v\ dv = -9 \int^{3}_{0}x dx$

$\dfrac{v^2}{2}|^0_v=-9\ \dfrac{x^2}{2}|^3_0$

$v^2=9\ \cdot\ 3^2$

$v=9\ m/s$ when x=0.

A ball is thrown up with velocity 20 m/s relative to ground in open lift moving up with speed 10 m/s at time of throwing and has an acceleration of $2 m/s^2$ upward, ball meets lift after

$h = u\left(t\right) + at^2$

$\implies bt^2 = 10t$

$\implies t = 5/3^3$

The height 'y ' and horizontal distance x of a projectile are given $y = 2t^2+t m$ and $x = 5 t$ m where t is in seconds then find the initial velocity of a projectile

$x = 5t$

$V_x = \frac {dx}[dt] = 5 m/s$

$y = 2t^2 + t$

$V_y = \frac {dy}{dt} =2 \times 2t +1$

$V_y = 4t + 1$

$\overrightarrow {V} = \overrightarrow {V_x} \hat {i} + V_Y \hat {j}$

$tan 45^0 = \frac {V_y}{V_x}$

$L = \frac {4 t + 1 }{5}$

$4t +1 = 5$

$4t = 4$

$t = 1 s$

The mass pf jupiter is 314 times that of earth and the diameter of jupiter is 11.35 times that of earth if 'g ' have a value of $9.8 ms^{-2}$ on the earth , what is its value on jupiter ?

Formula

$g = \frac { G \times M}{ R^2}$

$9.8 = \frac { G \times M}{ R^2}$

$mass J = 314 M$

$Radius R = 11.35 R$

$g'= \frac { G \times 314 \times M}{ 11.35 R \times 11.35 R}$

$= 2.43 \times \frac {GM}{R^2}$

$g '= 2.43 \times 9.8$

$= 23.90 m/s^2$

A projectile is fired at some angle $\theta$ with speed v such that it crosses two walls of equal height symmetrically as shown. The angle of projection with horizontal is

$\tan\theta = \dfrac{4}{3}$

$\theta = \tan^{-1}\left(\dfrac{4}{3}\right)$

$\theta = 53^{\circ}$

Two blocks of masses 10 g and 30 g are fastened to the ends of cord which passes over a frictionless pulley. The 10 g block rests on the table. What minimum force must be applied on the 10 g block to keep it on the table? What will be the tension in the cord under this condition? What will be the acceleration of the system and the tension in the cord when the force is withdrawn?

Tension when the force is withdrawn = 0.294 N
Acceleration = 4.9

$m/s^2$
Tension in the cord under the condition = 0.196 N
Force that must be applied = 0.147 N

If block is initially at rest find direction in which it will start moving on application of forces shown& also find acceleration

$fact = 20 +10 - 5$

$=25 N$

$fact - m \times a$

$25 = 2 \times a$

$a = 12. 5 m/s^2$

How many orbitals are possible for $n = 4$ ?

For n=4

The values for

$l$ can be from 0 to

$n-1$

From 0 to 3 i.e.

$l$ = 0, 1 ,2, 3

For

$l$ =0 i.e. s orbital (1)

$\Rightarrow$ 4s

$l=1$ i.e. p orbitals (3)

$\Rightarrow$

$p_x,p_y,p_z$

$l=2$ i.e. 4d orbitals (5 Orbitals)

$l=3$ there are 7 f orbitals

Total =16 orbitals

Total orbitals =

$n^2$ =

$4^2$ =16

Find acceleration vector of block.

$a = \dfrac{F}{m} = 8\hat{i} + 8\hat{j} + 18\hat{k}$

In the situation shown in figure the maximum value of horizontal force F to be applied on the block of mass 5 kg so that block A of mass A of mass 100 kg does not move upward the coefficient of static friction for contact surface of the block A is is 0.1 but the block B is placed on a smooth horizontal surface. If F =(774.6)x N. then find the value of x.

1878163_1901235_ans_525bc95803f640dba29f9c11a8db7900.jpg

A block of mass 'M' connected to an elastic spring of force constant K performs S.H.M. with amplitude 'A' . At the instant when the block is passing through its equilibrium position a small block of mass m falls on M oscillate together then what is the
(i) amplitude of oscillation?
(ii) the magnitude of frictional force on 'm' at an extreme position?

Let the initial condition be

$\lambda = A sin wt$

$v = \dfrac{dx}{dt} = A w cos wt$

It means V = Vmax = A w

m falls on M

Conserving momentum

$M Aw = (m+M) v'$

$\implies V' = \dfrac{M}{(m+M) }Aw$

A $60 \mathrm{kg}$ block slides along the top of a $100 \mathrm{kg}$ block. The lighter block has an acceleration of $3.6 \mathrm{m} / \mathrm{s}^{2}$ when a horizontal force $\mathrm{F}=350 \mathrm{N}$ is applied. Assuming there is no friction between the bottom $100 \mathrm{kg}$ block and the horizontal frictionless surface but there is friction between the blocks. Find the acceleration of the $100 \mathrm{kg}$ block during the time the $60 \mathrm{kg}$ block remains in contact.

1878169_1901229_ans_fb0125f5a7ae4f86baf34fc993a74ed6.jpg

Give the difference between the gravitational constant $(G)$ and acceleration due to gravity $(g)$ with reference of their value.

$g$ is a vector and

$g=9.8ms^{-1}$

$G$ is a scalar used in

$F=\frac{GM_1M_2}{R^2}$

$G=6.67*10^{-11}Nm^2kg$

A block of mass $4kg$ is placed at the point $A$ of a rough track. If slightly pushed towards right, it stops at the point $B$ of the track. Calculate the work done by the frictional force on the block during its transit from $A$ and $B$ .

$m=4kg,H=1m,h=0.8m$
Work done by frictional force = change in P.E

$w_{f}=mgh-mgH=mg(h-H)$

$w_{f}=4\times 10(0.8-1)=-4\times 2=-8$ J
1882851_1901364_ans_69811aa0e96d4e9aab5e6e05531a17cf.png

Solve :

Area of trapezium ABCD

$= \dfrac{1}{2} \times 10 \times (10 + 25)$

$= 35 \times 5 = 175\ m$

Area of trapezium OABE

$= 10 \times \dfrac{1}{2} (25 + 30)$

$= 55 \times 5$

$= 275\ m$

Total displacement

$= 175 + 275$

$= 450\ m$

1874260_1901632_ans_647205fa0bf446b192b854ae58f41838.png

What is the work done?

$A\to B$ &

$D\to C$

$\Delta v=0, w_{AB}= w_{BC}=0$

$w_{DC}=-P\Delta v=-3x(2-1)=-6\ atm \ l$

$w_{AD}^2 =-P\Delta v=-1\times (2-1)=-1\ atm\ l$

So, total work

$=|-6-1|l\ atm$

$=7\ l\ atm$

Obtain the relation between the magnitude of the linear (tangential) acceleration and angular acceleration in nonuniform circular motion.

angular acceleration is the rate of change of angular velocity

$\alpha =\dfrac{\Delta w}{\Delta t}$

linear acceleration is rate of change of linear velocity

$a=\dfrac{\Delta V}{\Delta t}$

we know

$v=wr$

So,

$a=\dfrac{r \Delta w}{\Delta t}$

$a= r \alpha$

angular acceleration is the rate of change of angular velocity

linear acceleration is rate of change of linear velocity.

The figure shows two masses $M$ and $m$ hanging from the ceiling, connected by springs with spring constant $k_{1}$ and $k_{2}$ . The mass $M$ is subjected to a driving force $F(t) = F_{0}\cos \omega t$ in the downward direction.
a) $(2\ points)$ Use Newton's second law to find the coupled equations of motion for the two masses.
b) $(3\ points)$ Assuming solutions of the form $x_{1} = A\cos \omega t$ and $x_{2} B\cos \omega t$ , find expressions for the steady-state amplitudes $A$ and $B$ in terms of $\omega, F_{0}, M, m, k_{1}$ and $k_{2}$ .
c) $(2\ points)$ For $\omega = \sqrt {k_{1}/M}$ show that the amplitude of vibration of mass $M$ will be zero if $k_{1}/ k_{2} = m/M$ .

At steady state

$k_{1}x_{1} = kx_{2} + m_{1}g+ F .... (1)$

$k_{1}x_{1} = mg + mg + F_{0}\cos \omega t$

$A_{1} = \dfrac {mg + mg + F_{0}\cos \omega t}{k_{1}}$

$A_{2} = \dfrac {mg}{k_{2}}$ .

1880893_1902321_ans_c2d90d5c59f74c1b9f02f7f9903e3650.jpg

An aeroplane has a run of $500\ m$ to take off from the runway. It starts from rest and moves with constant acceleration to cover the runway in $30\sec$ . What is the velocity of the aeroplane at the take off?

A particle tied to the end of a string oscillates along a circular arc in a vertical plane. The other end of the string is fixed at the centre of the circle. If the string has a breaking strength of twice the weight of the particle.
Find the acceleration of the particle at bottom and extreme.

$a_c$ at bottom

$= \frac{v_b^2}{R}$

$= \frac{gR}{R}$

$= g$

$a_c$ at top

$= \frac{v_T^2}{R}$

$= \frac{3gR}{R}$

$= 3g$

Given that $O_2O_1 = 0.5\,m, O_3O_1 = 0.3\,m.$ Find $m_1$ for which system is in equilibrium.

Frm figure we have

$O_2O_1 = 0.5\,m$

$O_3O_1 = 0.3\,m$

$cos\theta = \dfrac{O_3O_1}{O_2O_1} = \dfrac{3}{5}$

$\theta = 53^\circ$

$\Rightarrow T_2 sin\theta = 8\,N$

$\Rightarrow T_2 = \dfrac{8}{(4/5)} = 10\,N$

$\Rightarrow m_1g = T_2cos\theta \Rightarrow m_1 = \dfrac{3}{5} = 0.6\,kg$

1888953_1902615_ans_371dae3294074f2487f4edf37e1249aa.png

At a certain height above the surface of the earth, the gravitational acceleration is $90\%$ of its value at the earth's surface. Determine the height above the earth's surface .(radius of the earth is $6400km)$

$g=\dfrac{g_0}{\left(1+\dfrac{h}{R}\right)^2}$

$g=\dfrac{90}{100}\times g_0=\dfrac{g_0}{\left(1+\dfrac{h}{R}\right)^2}$

$g=\dfrac{g}{10}=\dfrac{1}{\left(1+\dfrac{h}{R}\right)^2}$

According to question

$1+\dfrac{h}{R}=\dfrac{\sqrt{10}}{3}$

$\dfrac{h}{R}=\dfrac{\sqrt{10}}{3}-1$

$h=R\left(\dfrac{\sqrt{10}}{3}-1\right)$

A body is projected from the surface of the earth so that is reaches at a height of $10R.g=9.8m/s^2.$ show that the velocity with which the body should be projected from the surface of earth is $V=14\sqrt{\dfrac{R}{11}}$ .

K.E of projection

$=P.E$ at height

$10R$

$\dfrac{1}{2}mv^2=mg10R$

$v^2=20gR$

$v=\sqrt{20gR}$

$g=\dfrac{Gm}{R^2}$

$$V=\sqrt{20\times \dfrac

{Gm}{R^2}\times R}$$

$V=\sqrt{20\dfrac{Gm}{R}}$

If $F=10mg$ , find tension in each sting.

As the system

$y$ in rest, so for block

$F\ net=0$

$T=T_A$

$T_A=T$

Now for

$IInd$ block,

$Fnet=0$

$\Rightarrow T_A=T_B$

$T_B=T$

Now for

$3rd$ block,

$T_B-mg=0$

$T_B=mg$

$T_A=T_B=mg$

Three blocks of masses $m_1 = 3m, m_2 = 2m$ and $m_3 = m$ are placed in contact on a horizontal frictionless surface as shown in Figure. A horizontal force F is applied to $m_1$ as shown. Match items in column I with those in column II.
Column I Column II
A) Net force acting on $m_2$ if F = 12 N p) 1 N
B) Net force acting on $m_2$ if F = 6 N q) 3 N
C) Net force acting on $m_3$ if F = 12 N r) 2 N
D) Net force acting on $m_3$ if F = 6 N
s) 4 N

$f_{2m} = 4$

$f_{2m} = 2$

$f_{m_3} = 2$

$f_{m_3} = 1$

Find total and binding energy of satellite of mass $800\ kg$ orbiting at $1800\ km$ above earth.
$(G = 6.67 \times 10^{-11} , R = 6400\ M = 6 \times 10^{24}$ all in SI units).

$TE = \dfrac{-GMm}{2(R + h)}$

$= \dfrac{-6.67 \times 10^{-11} \times 6 \times 10^{24} \times 800}{2 (6.4 + 1.8) \times 10^{6}}$

$= - 1.95 \times 10^{10} J$

$BE = - TH$

$= 1.93 \times 10^{10} J$

How do you calculate the number of photons?

Answer
You calculate the energy of a photon, and then you use the total energy to calculate the number of photons.

Number of photons

$= \dfrac{Total \ energy}{Energy \ of \ one \ photon}$

EXAMPLE
A common laser pointer produces 1.0 mW at a wavelength of 670 nm. Calculate the number of photons produced per millisecond.

Solution
Step 1. Calculate the energy of a photon.

$E=\dfrac {6.626 \times 10^{-34} \times 3\times10^8}{670\times10^{-9}}=2.965\times10^{-19}$

$= \dfrac{Total \ energy}{Energy \ of \ one \ photon} = 1.0 \times 10^{-3}J.s^{-1} \times \dfrac{1 photon}{2.965 \times 10^{-19}J} \times \dfrac{ 1s}{1000 ms} = 3.4 \times 10^{12}$ photons/ms

How to find the moment of inertia of a solid cylinder about transverse (perpendicular) axis passing through its center ?

Answer
It needs to be done in three steps.1. Stating Moment of Inertia of a infinitesimally thin Disk.2. Application of Perpendicular Axis and Parallel axis Theorems.3. Integrating over the length of the cylinder.But first of all let's state the problem.
Explanation:
Figure 1.Let us consider a cylinder of length L, Mass M, and Radius R placed so that z axis is along its central axis as in the figure.We know that its density

$\rho = \dfrac{Mass}{Volume} = \dfrac{M}{V}$
Let us consider that the cylinder is made up of infinitesimally thin disks each of thickness dz. If dm is the mass of one such disk, then

$dm = \rho \times Volume of dist$

$dm = \dfrac{M}{V} \times (\pi R^{2}dz)$ since
V = Areal of circular face

$\times length = \pi R^{2}L$ . we obtain.

$dm = \dfrac{M}{\pi R^{2}L} \times (\pi R^{2}dz)$

$dm = \dfrac{M}{L}dz$
Step 1.
We know that moment of inertia of a circular disk of mass m and of radius R about its central axis is is same as for a cylinder of mass M and radius R and is given by the equation

$I_{z} = \dfrac{1}{2}mR^{2}$ In our case

$dI_{z} = \dfrac{1}{2}dmR^{2}$
Step 2.
Observe from figure 2, that this moment of inertia has been calculated about z axis. In the problem we are required to find moment of inertia about transverse (perpendicular) axis passing through its center. Knowing that the desired axis of rotation is transverse, therefore we need to apply perpendicular axis theorem which states:
The moment of inertia about an axis which is perpendicular to the plane contained by the remaining two axes is the sum of the moments of inertia about these two perpendicular axes, through the same point in the plane of the object.It follows that

$dI_{z} = dI_{x} + dI_{y}$
Also from symmetry we see that moment of inertia about x axis must be same as moment of inertia about y axis.

$dI_{z} = \dfrac{dI_{z}}{2}$ Substituting

$I_{z}$ from (2), we get

$dI_{z} = \dfrac{1}{2} \times \dfrac{1}{2}dmR^{2}$

$dI_{x} = \dfrac{1}{4}dmR^{2}$
Let the infinitesimal disk be located at a distance z from the origin which coincides with the center of mass.
Now we make use of the parallel axis theorem about the x axis which states:
The moment of inertia about any axis parallel to that axis through the center of mass is given by

Write down some properties of liquids.

Properties of liquids:

Particles are held by bonds weaker than solids' yet stronger than a gas.
It takes the shape of its container.
It can evaporate to create a gas.
It can be frozen to create a solid

What do you mean by ionization energy?

As you know, the ionization energy is the energy needed to remove 1 mole of the outermost electrons from 1 mole of gaseous atoms in the ground state.

$X_{(g)} + I.E. \rightarrow X_{(g)}^+ + e^-$

Hydrogen has an ionization energy of

$1312 \ kJ \ mol^{-1}$ , which means that in order to remove 1 mole of valence electrons from 1 mole of gaseous hydrogen atoms in the ground state, you need to provide

$1312 \ kJ$ of energy.

Is $SO_2$ a liquid, a solid, or gas at room temperature ?

$SO_2$ is a gas at room temperature.

According to the quantum $-$ mechanical model transition would produce light with the longar wavelength : $2p \rightarrow 1s$ or $3p \rightarrow 1s$ ?

How many atoms in a mass of ONE MOLE of magnesium nitrate?

An isotope of iron has $28$ neutrons. If the atomic mass of the isotope is $54$ , how many protons does it have?

The iron atom is characterized by an atomic number,

$Z=26$ . That is every iron nucleus contains 26 protons, 26, massive, positively charged, nuclear particles, by definition.

Of course, the nucleus also contains 28 neutrons, 28 neutral, massive nuclear particles, to give the

$^{54}Fe$ isotope, which is about 4% abundant, check on this.

The number of protons, the number of massive, positively charged, nuclear particles, ALWAYS defines the identity of the element.

An unstretched spring has a length of 10 centimeters. When the spring is stretched by a force of 16 newtons, its length is increased to 18 centimeters. What is the spring constant of this spring?

Force on a spring is given by

$F= - k \Delta x$

$k= \dfrac{- F}{\Delta x} = \dfrac{-(-16 \ N)}{18 \ cm - 10 \ cm} = 2 \ N/ cm = 200 \ N/m$

A stone is thrown vertically upwards and it reaches a maximum height of $30m$ . What was its initial velocity?

$24.248 ms^{-1}$
Explanation:
A stone is thrown vertically upwards. So that the only force acting on teh stone is the Gravitational force. we consider teh upward direction as positive

$y-axis$ . Then there is no motion is along

$x-axis$ . And the acceleration of the stone is

$-g$
where

$g$ is acceleration due to gravity
the negative indicates that the acceleration is along negative

$y-axis$ (towards downwards direction).
The equation of motion connecting the velocities and the displacement of the particle is given by

$v^2=u^2+2a.S$
Here

$v$ is the final velocity,

$u$ is the initial velocity (velocity with which the stone is thrown)

$a$ is the acceleration, here (

$-g$ acceleration due to gravity),

$S$ is the displacement, here (Maximum Height reached by the stone)
At the maximum height velocity becomes zero

$(v=0)$
SUBSTITUTE VALUES IN THE EQUATION

$.0=u^2-2.g.H$

$u^2=2.g.H$

$u=\sqrt{2.g.H}$

$u\sqrt{2 \times 9.8 \times 30}$

$u=24.248\ ms^{-1}$
Therefore the velocity with which the stone thrown i s

$24.248 ms^{-1}$

What is the de Broglie wavelength for an electron that travels at $90\%$ the speed of light?

Explanation:
First we need to know the formula. It is ,

$\lambda =\dfrac{h}{mv}$ we know speed of light is

$3.00 \times 10^8 m/s$ , then we need to find what

$90\%$ of that is.

To find that we do this,

$0.90c=3.00 \times 10^8 m/s \times 0.90$

$=2.7 \times 10^8 m/s$

$\lambda=\dfrac{h}{mv}$

$\lambda=wavelength$

$h=$ planck's constant, which is

$6.63 \times 10^{-34} m^2 . kg/s$

$m=$ mass of electron, which

$9.11 \times 10^{-31} kg$

$v=$ velocity

$\lambda =\dfrac{6.63 \times 10^{-34} kg. m^2/s}{(9.11 \times 10^{-31} kg) \times (2.7 \times 10^8 m/s)}$

$\Rightarrow \lambda= 2.70 \times 10^{-12} m$

What is the molar mass of gold metal?

Molar mass of Au = 197 gm/mol

If two people are pulling on the same rope at 10 N of force, what is the tension on the rope?

$10\,N$

Explanation:

This is an application of Newton's Third Law: "For every action there is an equal and opposite reaction." The only way someone on one end of the string can exert a force of 10 N on the other end of the string is of the other end of the string exerts a 10 N force in the opposite direction.

An atom of an element has 9 electrons 9 protons and 10 neutrons. Then what is the atomic number?

Answer
The atomic number is 9.
Explanation:
The atomic number of an element is the number of protons in one atom of that element. Thus, if an atom of an element has nine protons, then its atomic number is also nine.

Are protons positive, neutral or negative?

Answer
Positive.
Explanation:
Protons have a positive charge of 1e, which is equal to about

$1.602 . 10 ^{-19} C$

A compound has a molar mass of $142.00g/mol$ and the percent composition is $50.7\%$ carbon, $4.2\%$ hydrogen and $45.1\%$ oxygen. What is the molecular formula ?

Every time a problem provides you with a compound's molar mass, you can use a little trick to help you determine the compound's molecular mass faster.
More specifically, instead of determining the empirical formula first, then using the molar mass to get the molecular formula, you can skip the empirical formula altogether.

As you know, a compound's molar mass tells you what the mass of one mole of that substance is. In this case, one mole of your compound has a mass of

$142.00\ g$ .
This means that if you pick a sample of

$142.00\ g$ of this compound, you can use its percent composition to get the exact number of moles of each element you get per mole of compound.

So, you know that this compound's percent composition is as follows

$carbon \rightarrow 50.7\ %$

$hydrogen \rightarrow 4.2\%$

$oxygen \rightarrow 45.1\ %$

This means that it contains

$142.00\ g\ compound\times \dfrac{50.7\ g\ C}{100\ g\ compound} = 71.994\ g\ C$

$142.00\ g\ compound\times \dfrac{4.2\ g\ H}{100\ g\ compound} = 5.964\ g\ H$

$142.00\ g\ compound\times \dfrac{45.1\ g\ O}{100\ g\ compound} = 64.042\ g\ O$

Now, all you have to do is use the molar masses of these three elements to determine how many moles of each element you get in one mole of your compound.

For $C:\ 71.994\ g\times \dfrac{1\ mole\ C}{12.011\ g} = 5.994 \approx 6$

For $H:\ 5.964\ g\times \dfrac{1\ mole\ H}{1.00794\ g} = 5.92 \approx 6$

For $O:\ 64.042\ g\times \dfrac{1\ mole\ O}{15.994\ g} = 4.003 \approx 4$

Therefore, the molecular formula of the compound will be $C_6H_6O_4$ .

A train leaves Chicago traveling west at $60$ miles per hour. An hour later, at $12$ noon, another train leaves Chicago traveling east at $80$ miles per hour. When are the two trains the same distance from Chicago?

Answer
3:00 p.m.
Explanation:
Since we are just talking about distance from Chicago it doesn't matter what direction they are going.
What does matter is the speed of each train and the head start the first train had.
The first train's distance can be represented with the equation:
60 + 60 x because it has an hour head start where it travelled 60 miles in that time.
The second train's distance can be represented with the equation:
80 x because each hour it travels 80 miles.
In both equations x represents the number of hours.
If we set these two equations equal to each other we get:
60 + 60 x = 80 x
Combine like terms:
60 = 20 x
Divide both sides by 20:
x = 3
So at 3:00 p.m. the two trains will both be the same distance from Chicago (240 miles).

A total of 640 joules of work is done on a 50.-kilogram object as it is moved 8.0 meters across a level floor by the application of a horizontal force. What is the magnitude of the horizontal force applied to the object?

Answer
The force is 80 N.
Explanation:
Work is the product of force, displacement, and the cosine of the angle between them. In this case, it says that the work done is 640 J, displacement is 8.0 m. It also says that the force is horizontal, so the angle between force and displacement is zero.
Therefore,640 J = F8.0 mcos 0 = F8.0 m
Solving for F,

$F = \dfrac{640 J}{8.0} = 80 N$

A $4\ kg$ ball has a momentum of $12\ kg m/s$ . What is the ball's speed?

$v=3 \dfrac{m}{s}$
Explanation:

$m=4\ kg$ mass of ball

$v=?$ velocity of ball

$P=12 kg \dfrac{m}{s}$ momentum of ball
an object's momentum is calculated using formula

$P=m.v$

$12=4.v$

$v=\dfrac{12}{4}$

$v=3 \dfrac{m}{s}$

A compound is found to contain $40.0\%$ carbon, $6.7\%$ hydrogen and $53.5\%$ oxygen. Its molecular mass is $60\ g/mol$ . What is its molecular formula ?

$100\ g$ of the unknown compound, the ratio of

$C\ ,\ H$ and

$O$ is :

$\dfrac{40.0 \ g}{12.011 \ g \ mol^{-1}} C : \dfrac{6.7 \ g}{1.00794 \ g \ mol^{-1}} H:\dfrac{53.5 \ g}{ 16.00 \ g \ mol^{-1}}O$ .

We divide through to get

$C : H : O= 3.33 : 6.65 : 3.34$ .

When we divide each elemental ratio by the LOWEST number, we get an empirical formula of

$CH_{2}O$ . i.e, near enough to WHOLE numbers.

The empirical formula is

$CH_{2}O$ , and the molecular formula will be some multiple of this.
Now the molecular formula is always a multiple of the empirical formula; i.e,

$(EF)_{n}= MF$ .

$60.0 \ g \ mol^{-1} = n \times (12.011 + 2 \times 1.00794+16.00)g \ mol^{-1}$ .
Clearly

$n=2$ , and the molecular formula is

$2 \times (CH_{2}O)= C_{2}H_{4}O_{2}$ .

Therefore, the molecular formula of the given compound is

$C_{2}H_{4}O_{2}$ .

A sample of ammonium nitrate having a mass of 3.88 grams of water is dissolved in 60.0 g of water. The temperature of the water decreases from $23.0 ^{\circ} C$ to $18.4 ^{\circ} C$ What is the molar enthalpy of dissolution for ammonium nitrate ?

Total mass of the solution = 60+3.88 g = 63.88 g

$C_p \ of \ water = 4.18 \ J/g ^oC$

$q_{solution} = -63.88 \times 4.18 \times (23-18.4) =-1228.28 \ J$

$q_{rxn} = -q_{solution} = 1228.28 \ J$

$Moles \ of \ NH_4NO_3 = \dfrac{3.88}{80} = 0.0485$

$Enthalpy = \Delta H = \dfrac{1228.28 \ J}{0.0485 \ mol } =25325.36 \ J/mol$

A transverse wave is described by the equation, $y = y_{0}\sin 2 \pi \left ( ft - \frac{x}{\lambda } \right )$ . The maximum velocity of a particle will be $4$ times the velocity of wave provided if $\lambda$ equals to?

A transverse wave is described by equation :

$y=y_0 \sin 2 \pi \left( ft - \dfrac{x}{\lambda} \right)$

The maximum velocity of a particle will be :

$\nu_{max}= amplitude \times 2 \pi f =y_0 \times 2 \pi f$

Velocity of wave :

$\nu_{wave}= \lambda \times f$

By the condition of the problem

$\nu_{max}= 4 \times \nu_{wave}$

$\Rightarrow y_0 \times 2 \pi f =4 \times \lambda \times f$

$\Rightarrow \lambda= \dfrac{\pi y_0}{2}$

During takeoff, an aeroplane climbs with a speed of $163 m/s$ at $31^{\circ}$ above the horizontal. The speed + direction of the aeroplane constitutes a vector quantity called velocity. The sun shines overhead. How fast is the shadow of the plane moving along the ground?

$\cos\left(31^{\circ}\right)$ .

$163 m/s = 139.7 m/s$
You can look at the vector as having a vertical and a horizontal velocity which has a vector sum of magnitude

$163m/s$ at angle

$31^{\circ}.$
We can use trigonometry to see that the horizontal speed is the cosine of the angle times the total speed. The horizontal speed of the aeroplane is the horizontal speed of the shadow.

How do you calculate displacement on a velocity vs.time graph?

Area under the curve - broken in two parts - the area above the and the area below the time axis. the sum (including a negative sign for area below) would give displacement.
Explanation:
Area below the curve is distance in a velocity time graph gives the distance, However, to find the displacement we need to break the area into two parts.

$(i)$ the area between the curve and above the time axis and

$(ii)$ the area between the curve and below the time axis
The area below the time axis would be negative as

$v$ is negative.
Adding the two gives the displacement.

Are protons and neutrons smaller than electrons?

Electrons are actually much smaller than protons and neutrons.

Protons and neutrons have masses which are equal to 1 atomic mass unit. Technically they vary from each other slightly, but most classroom applications don't require you to use this information.

The mass of an electron is about 1/1800 the mass of a proton. This is why the mass of electrons is ignored when you are determining the mass of an atom.

The volume occupied by the nucleus (protons and neutrons) is much, much smaller than the volume occupied by the electron cloud which is made up of the electrons of an atom and mostly empty space (since electrons are so tiny). This means that protons are neutrons are packed together into a small space and that electrons are very spread out.

How do you calculate the number of molecules in a substance?

$\text {No. of molecules} = \cfrac {\text {Weight of substance}}{\text {Molar Mass}} \times N_A$

What do scientists call anything that takes up space and has mass?

Matter is anything that occupies space space(volume) and has mass.

Matter exist in several states: Solid, liquid and gas.

How did James Chadwick contribute to the atomic theory?

Answer
James Chadwick played a vital role in the atomic theory, as he discovered the Neutron in atoms.
Neutrons are located in the center of an atom, in the nucleus along with the protons. They have neither a positive nor negative charge, but contribute the the atomic weight with the same effect as a proton.
Chadwick discovered this subatomic particle by using a neutron chamber in his experiments. Here is a link to a visual design of his experiment

How is Thomson's model of an atom different from Dalton's model?

John Dalton proposed that all matter is composed of very small things which he called atoms. This was not a completely new concept as the ancient Greeks (notably Democritus) had proposed that all matter is composed of small, indivisible (cannot be divided) objects. He thought atoms to be literally 'atomos' meaning 'uncuttable'

Later J. J. Thomson using his Cathode ray tube experimented and found out that atoms were made up of differently charged particles. This he called the plum pudding model. The Plum Pudding Model is a model of atomic structure proposed by J.J. Thomson in the late 19th century.

Thomson had discovered that atoms are composite objects, made of pieces with a positive and negative charge and that the negatively charged electrons within the atom were very small compared to the entire atom. He, therefore, proposed that atoms have a structure similar to a plum pudding, with tiny, negatively charged electrons embedded in a positively charged substrate. This was later shown to be incorrect.

The formula for aluminum sulfate is $Al_{2}(SO_{4} )_{3}$ . What is the molar mass of aluminum sulfate to the nearest gram?

$M_{M Al} = 26.981538 g mol^{-1}$

$M_{M S} = 32.065 g mol^{-1}$

$M_{M O}=15.9994 g mol^{_1}$
In order to find the molar mass of aluminium sulfate, you will need to do

$2 \times M_{Al} + 3 \times M_{S} + 12 \times M_{O}$
Rounded to the nearest gram, the answer will be

$M_{M Al_{2}} (SO_{4})_{3} = 342 g mol^{-1}$

What are the highest frequency and lowest frequency parts of the EM spectrum?

Answer
See below...
Explanation:
Radio waves usually have the lowest frequency, while Gamma rays have the highest frequency. And as you can see the trend is opposite for wavelength, as frequency and wavelength are inversely related.
1906407_1919363_ans_4a8407a5fa124248aca6a458d00cac3c.PNG

1906407_1919363_ans_4a8407a5fa124248aca6a458d00cac3c.PNG

What is differentiating electron?

The differentiating electron is the last electron that is added when you are building up the electron configuration of an atom.

It is the electron that differentiates the atom from the one just before it in the Periodic Table.

For example, the electron configuration of gallium is

$[Ar] 4s^2 3d^{10}\ 4p^1$ .

The element just before gallium in the Periodic Table is zinc.

Its electron configuration is

$[Ar]\ 4s^2 3d^{10}$ .

Thus, the differentiating electron in gallium is the

$4p$ electron.

Why does air have pressure?

Air pressure exists because air has mass.

This column of air has a mass of about $10,000 \mathrm{Kg}$ . This exerts a force of about $10,000 \mathrm{N}$ due to its weight under gravity.

The unit of air pressure is the Pascal or Pa.

$1 P a=1 \dfrac{N}{m^{2}}$

The average air pressure at sea level is $101,325 \mathrm{Pa}$ .

Air in a container, like any gas, also has pressure due to the air molecules bouncing off the container walls.

How did Ptolemy's model account for retrograde motion?

Answer
Ptolemy included epicycles in his orbits.
Explanation:
Ptolomy's model of the solar system was geocentric, where the sun, moon, planets, and stars all orbit the earth in perfectly circular orbits. The problem with perfectly circular orbit around the Earth is that they do not explain the occasional backward motion, or retrograde motion, of the planets
The Greeks insisted that the motion of the planets be perfectly circular. Ptolemy modeled the planets making small circles around a point that orbited the Earth. These smaller circles were called epicycles, and they allowed the planets to move backward relative to the background stars.
Ptolemy's model took epicycles even further, using them to explain the brightening and dimming of the planets as well, by having epicycles attached to epicycles. While these epicycles did not perfectly explain the motion of the planets, it was the most accurate model until Kepler's laws simplified things.
1906422_1919376_ans_d89542fcf810479091fc27c3c331c268.PNG

1906422_1919376_ans_d89542fcf810479091fc27c3c331c268.PNG

Can the magnitude of a vector be negative?

No, The sign in a vector indicates it's direction, but the magnitude is always positive (or zero).

A window washer pulls herself upward using the Bucket-pulley apparatus. The mass if the person plus the Bucket is 58 kg. How hard must she pull downward to raise herself slowly at a constant speed?

$Answer$

$F_{pull} = 570 N$

$Explanation :$

If she is rising at a constant speed, then the net force on her is 0.
The only two forces acting on her and the bucket are the (1) weight, and the (2) upward force exerted by the rope on the woman (opposite in direction and equal in magnitude to the downward pulling force she exerts on the rope).
Thus

$F_{pull} = mg$
We know

$m = 58 kg$

$g = 9.81 m / s^2$
Therefore

$F_{pull} = (58kg) (9.81 m / s^2) = 570 N$ rounded to 2 significant figures.

1919994_1924996_ans_b02c825aca7740b1b01369f19d96ad03.PNG

How do you solve for P in PV = nRT?

$P = \dfrac{nRT}{V}$

Calculate energy of transition from n=3 to n=What region of electromagnetic spectrum is this radiation found?

Answer
Since you did not specify, I'm going to assume you mean, hydrogen with a single electron.

$E_{ph} = 1.89 e V$

$\lambda = 655.2$ this is in the optical (it's a red-pink)
Explanation :
The formula

$E_{ph} = hc R \left ( \left ( \dfrac{1}{n_{lower}} \right )^2 - \left ( \dfrac{1}{n_{upper}} \right )^2 \right )$

$hcR = 13.6 e V$

$E_{ph} = 13.6 eV \left ( \left ( \dfrac{1}{2} \right )^2 - \left ( \dfrac{1}{3} \right )^2 \right )$

$E_{ph} = 13.6 eV \left ( \dfrac{1}{4} - \dfrac{1}{9} \right )$

$E_{ph} = 13.6 eV \left ( \dfrac{5}{36} \right ) = 1.89 e V$

$E_{ph} = h \left ( \dfrac{c}{\lambda} \right ) = hcR \dfrac{5}{36}$

$\dfrac{1}{\lambda} = R \dfrac{5}{36}$

$\lambda = \dfrac{1}{R} \dfrac{36}{5}$

$\dfrac{1}{R}$ is a constant and 91 nm.

$\lambda = 655.2$
this is in the optical (it's a red-pink)

What is the shape of f-orbital???

2033125_1925608_ans_83bc3122ac60410abb81fbe5fedd8f1e.png

What number of electrons would carry a total charge of one coulomb?

The number of electrons per coulomb:

$\dfrac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}$ electrons per coulomb.

A $4000\ kg$ truck is parked on a $7.0^{\circ}$ slope. How big is the friction force on the truck?

$\vec{f}_{s} \approx 4.8 \times 10^{3}\ N$
Explanation :
Here is a basic force diagram for an object on an inclined plane:

Note that the angle is certainly not to scale.
Note that I will define down the ramp as the positive direction.

We can determine the frictional force acting on the truck with statements of the parallel and perpendicular components of the net force acting on the truck.

$\displaystyle \sum F_{x} = \vec{F}_{gx} - \vec{f}_{s} = m \vec{a}_{x}$

$\displaystyle \sum F_{y} = \vec{n} - \vec{F}_{gy} = m \vec{a}_{y}$

To prevent the truck from rolling down the incline, we want a horizontal acceleration of zero. The vertical acceleration should also be zero, as we certainly don't want the truck to be moving in the vertical direction. This leaves us with a net force of zero on the truck. Note this is static friction.

$\displaystyle F_{x} = \vec{F}_{gx} - \vec{f}_{s} = 0$

$\displaystyle F_{y} = \vec{n} - \vec{F}_{gy} = 0$

We can solve for the force of friction using the parallel sum of forces statement. The perpendicular forces are not necessary.

$\vec{F}_{gx} - \vec{f}_{s} = 0$

$\Rightarrow \vec{F}_{gx} = \vec{f}_{s}$

The parallel component of the force of gravity is given as $$mg \sin(\theta) $$ for this situation, which we can see from the force diagram. I will explain this at the end if it is unclear.

$\Rightarrow \vec{f}_{s} = mg \sin (\theta)$

Using our known values...

$\Rightarrow \vec{f}_{s} = (4000\ kg) \left (9.8 \dfrac{m}{s^2} \right ) \sin (7^{\circ})$

$\Rightarrow \vec{f}_{s} \approx 4777\ N$

You might write $4,8 \times 10^{3}\ N$ where significant figures are concerned.

If no external forces act on a moving object, it will? a) move slower and slower until it finally stops. b) come to an abrupt halt. c) continue moving at the same speed. d) none of the above

What is the difference between $Fg = mg$ and Universal Gravitation?

Answer
If an object of mass m is on the surface of the earth of mass M and radius R then by the law of universal gravitation the force of gravity on the object is given by

$F_{g} = \dfrac{GmM}{R^2} ...[1]$
where G is the universal gravitational constant.
Again acceleration due to gravity (g) is the acceleration with which any object freely falls under the force of gravity. towards the centre of the earth.
Now if the mass of the body be m then by Newton's laws of motion the gravitational pull on the object will be given by

$F_{g} = mass \times acceleration = m \times g ....[2]$
So the origin of this equation is Newton's laws of motion. Here only acceleration (g)is originated from the gravitational force.
Comparing equation [1] and [2] we can write

$mg = \dfrac{GmM}{R^2}$

$\Rightarrow g = \dfrac{GM}{R^2}$
This equation shows how the acceleration due to gravity (g) is related with universal gravitational constant (G) as well as mass(M) and radius (R)of the earth

What is the de Broglie wavelength of an electron in the ground state of a hydrogen atom ?

The formula for the de Broglie wavelength

$\lambda$ is

$\boxed{\lambda = \dfrac{h}{mv}}$

where

$h$ = Planck's constant

$m$ = the mass of the electron

$v$ = the speed of the electron.

Calculate the speed of the electron

The energy

$E$ of a hydrogen electron in an orbit is

$E = - \dfrac{R_{H}}{n^2}$

where

$R_{H}$ = the Rydberg energy constant 2.180 \times 10^{-18}\ lJ) $$

Since

$n = 1$

$E_{1} = -2.180 \times 10^{-18}\ J$

$KE = -E_{1} = 2.180 \times 10^{-18}\ J$

$KE = \dfrac{1}{2} mv^{2}$

$v = \sqrt{\dfrac{2KE}{m}} = \sqrt{\dfrac{2 \times 2.180 \times 10^{-18}\ J}{9.11 \times 10^{-31}\ kg} \times \dfrac{1\ kg \cdot m^{2}s^{-2}}{1\ J}} = \sqrt{4.786 \times 10^{12}\ m^{2}\ s^{-2}}$

$= 2.188 \times 10^{6}\ m \cdot s^{-1}$

Calculate the de Broglie wavelength

$\lambda = \dfrac{h}{mv} = \dfrac{6.626 \times 10^{-34}\ J \cdot s}{9.109 \times 10^{-31}\ kg \times 2.188 \times 10^{6}\ m \cdot s^{-1}} \times \dfrac{1\ kg \cdot m^{2} \cdot s^{2}}{1\ J} = 3.325 \times 10^{-10}\ m = 332.5\ pm$

Ans-

$332.5\ pm$

What refers to how large or strong is the force.

Explanation: Magnitude refers to the size or strength of the force.The mass of an object refers to the amount of matter that is contained by the object; the weight of an object is the force of gravity acting upon that object.

A body is thrown up in a lift with a velocity u' relative to the lift and the time of flight is found to be t'. the acceleration with which the lift is moving up will be a

Let the acceleration of the lift be a in upward direction.

Acceleration of the body w.r.t ground =g (downward)

Thus Acceleration of the body w.r.t lift

$a_b=g+a$

In the lift frame :

$S=ut+\dfrac{1}{2}a_bt^2$

where t is the time of flight

∴

$0=ut−\dfrac{(g+a)}{2}t^2$

⟹

$a=\dfrac{2u−tg}{t}$

1953854_1936168_ans_8bbf77d159c5492c850a46de2148039a.png

Manav drives from his home to the nearest metro station at an average speed of $40$ kmph. From the station, he boards a train that takes him to his office at a speed of $80$ kmph. The entire distance covered by him is $52$ km and the entire journey took him $1$ hour. Find the distance between the station from where he boards the train and his office?

The distance from home to station is

29 km and the distance from station to office is

23 km.
Step-by-step explanation:
Define

x:
Let the journey by car be

x km
The journey by train is

(52−x)km
Find the time taken to drive to the station:
Time Taken = Distance

÷ Speed
Time Taken =

x/40 hour
Find the time taken for the journey on the train:
Time Taken = Distance

÷ Speed
Time Taken =

(52−x)/80
Solve

x:
The total time taken is

1 hour

x/40+(52−x)/80=1

(2x+52−x)/80=1

2x+52−x=80

x+52=80

x=29km
Find the distance:
Distance from home to station =

x =

29 km
Distance from station to office =

52−x =

52−29 =

23 km
Answer: The distance from home to station is

29 km and the distance from station to office is

23 km.

How do vectors work in physics?

Vectors are lines that represent both magnitude (size) and direction. The length of the vector is drawn to scale, to communicate the size of the measurement; the head of the arrow points in the direction in which the measurement is traveling.
Vectors can be used in a number of ways. Most commonly, they are used to represent displacement or velocity of a moving object, but they can also be used to represent force.
If an object moves in more than one direction subsequently, or if more than one force acts upon an object concurrently, vectors can be added to find a resultant displacement or resultant force. This can be calculated using either a scale diagram and measuring the resultant, or by use of mathematics.
1914188_1908392_ans_910f6d18d5a64d9992182cb1269b451b.png

1914188_1908392_ans_910f6d18d5a64d9992182cb1269b451b.png

What is the formula for average force?

The force applied by a body that’s travelling at a definite velocity (rate of speed) for a definite period of time is the average force. Force is a vector quantity that has both magnitude and direction. The word ‘average’ is made use of to specify that this velocity is not an accurately measured or ‘instantaneous’ velocity. Therefore, the mass of the body multiplied by the average velocity over the definite time is equivalent to average force. Average force is a vector quantity that has both magnitude and direction.

For a particular interval of time t, the force is described as the frequency of change of momentum. It is hard to compute the rate of change if the time interval is minor. There the term, average force makes an entrance.

Over a period of intervals,

$(\Delta t)$ the rate of change of momentum is Average Force. It is given by

$F =\dfrac{ m (v_f – v_i)}{\Delta t}$

Where,

the mass of the body is

$m,$

the final momentum is

$v_f,$

the initial momentum is

$v_i,$

the change in time is

$\Delta t.$

The Average Force Formula aids one in getting the rate of change of momentum for any number of time intervals

$(\Delta t).$ Expressed in Newton

$(N).$

Column I	Column II
A) Net force acting on $m_2$ if F = 12 N	p) 1 N
B) Net force acting on $m_2$ if F = 6 N	q) 3 N
C) Net force acting on $m_3$ if F = 12 N	r) 2 N
D) Net force acting on $m_3$ if F = 6 N	s) 4 N

Basic Science - Class 5 General Knowledge - Extra Questions

Why do the glaciers move?

Atom of which element has two protons and two electrons ?

How do you write an equation for the reaction between zinc metal and hydrochloric acid (aqueous) producing zinc chloride (aqueous) and hydrogen gas?

A man descending by parachute does it slowly. Why?

A tennis ball is thrown up and reaches a height of 4.05m4.05 m before coming down. What was its initial velocity? How much total time will it take to come down? Assume g=10 m/s2g = 10\ m/s^{2}.

Consider the situation shown in figure. The is released from rest and the block of mass 1.0 kg1.0\ kg is found to have a speed 0.3 m/s0.3\ m/s after it has descended through a distance of 1 m1\ m. Find the coefficient of kinetic friction between the block and the table.

A cyclist speeding at 18 km/h18\ km/h on a level road takes a sharp circular turn of radius 3 m3\ m without reducing the speed. The co-efficient of static friction between the tyres and the road is 0.10.1. Will the cyclist slip while taking the turn?

Prove that the angular momentum of a satellite of mass MsM_{s} revolving round the earth of mass MeM_{e} in an orbit of radius rr is equal to [GMeM2sr]1/2[GM_{e}M_{s}^{2}r]^{1/2}.

In the x−tx-t graph below, indicate the time when the average velocity is equal to the instance velocity.

A car of mass mm is coasting on a horizontal surface. There is no friction between the car and the surface, but a force of air resistance acts on the car of magnitude Fdrag(v)=kv2F_{drag}(v) =kv^2. The initial velocity of the car is v0v_0.

Diagram shows a velocity-times graph for a car starting from rest. The graph has three sections AB,BCAB,BC and CDCD.In which section, car has a zero acceleration.

A car with a gun mounted on it is kept on horizontal friction less surface total mass of car , gun and shell is 0 kg mass of each shell is 1 kg if shell is fired horizontally with relative velocity 100 m/ sec with respect to gun what is the recoil speed of car after second shot in nearest integer ?

A particle if given a certain velocity v at point P as shown on a hemispherical smooth surface find the value of v (in m/s) such that when particle reaches Q , the normal reaction of surface becomes equal to particle's weight [R=1.6m,g=10m/s2] [ R = 1.6 m , g = 10 m/s^2 ]

Diagram shows a velocity-times graph for a car starting from rest. The graph has three sections AB,BCAB,BC and CDCD.Is the magnitude of acceleration higher or lower than that of retardation? Give Reason.

The position of a particle along the x−x-axis is given in centimeter by x=9.75+1.50t3x=9.75+1.50 t^3, where tt is in second. Consider the time interval t=2 st=2\ s to t=3 st=3\ s and calculate the average velocity.

A person moves through a distance of 20 m20\ m due to east, then moves through a distance 20 m20\ m due north, then turns west and covers a distance of 10 m10\ m and then turns south and covers a distance of 10 m10\ m. Find the displacement.