Basic Science - Class 5 General Knowledge - Extra Questions

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.

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 required reaction willbe:

It is due to variation in pressure of the atmosphere.

$$H_{max} = \dfrac {u^{2}\sin^{2} 90^{\circ}}{2g}$$
$$4.05 = \dfrac {u^{2}\times 1}{2\times 10}$$
$$u^{2} = \dfrac {4.05}{20} = 0.2025$$
$$u = \sqrt {0.025} = 0.45\ m/s$$
$$T = \dfrac {24\sin 90^{\circ}}{g} = \dfrac {2\times 0.45}{10}$$
$$= 0.09\sec$$.

$$T-f_{k}=4\times (2a)...(i)$$
$$10-2T=1\times a....(ii)$$
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 $$u_{k}=0.23$$

$$v=18\ km/h=18\times \dfrac{5}{18}=5\ m/sec$$
$$R=3\ m$$
$$f=\mu mg$$
$$=0.1\times m\times 10$$
$$f=m$$  ....(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

$$F=\dfrac{GM_{e}M_{s}}{r^{2}}=\dfrac{MS_{2}}{r}$$
($$G$$. force)
(centripital force)
$$=\dfrac{GM_{e}}{r}=v^{2}$$
$$v=\sqrt{\dfrac{GM_{e}}{r}}$$
Angular momentum $$=M_{S}\cup r$$
$$=MS\sqrt{\dfrac{GM_{e}}{r}}\times r$$
$$=\sqrt{GM_{e}M_{s}^{2}r}$$

nstantaneous $$\Rightarrow \dfrac{dx}{dt}=\dfrac{x_1-X_6}{t_1-0}$$
$$avg=\dfrac{displacement }{time}$$
$$=\dfrac{X_1-X_0}{t_1-0}$$ 

Let at any instant reativity of car is $$v$$
$$\displaystyle \int =kv^2$$
$$ma =kv^2$$
$$m \dfrac {dv}{dt}=kv^2$$
$$\displaystyle \int_{v_0}^{v\ dv } = \displaystyle \int_{t=0}^{t }\dfrac km v^2\ dt$$
$$\displaystyle \int_{v_0}^v  \dfrac {dv}{dt} = \dfrac km \displaystyle \int_0^t dt$$ 
$$\left[-\dfrac {1}{v}\right]_{v_0}^v =\dfrac {k}{m}t$$
$$-\left(\dfrac {1}{v}-\dfrac {1}{v_0}\right)=\dfrac {k}{m}t$$
$$\dfrac {1}{v}-\dfrac {1}{v_0}=-\dfrac {k}{m}t$$
$$\dfrac {1}{v}=\dfrac {1}{v_0}-\dfrac {k}{m}t =\dfrac {m-v_0 kt}{v_0 m}$$
$$v=\dfrac {mv_0}{m-v_0 kt}=\left[\dfrac {v_0}{1-\dfrac {v_0}{m}kt}\right]$$

$$ N_1 cos \beta  = mg $$
$$ N_ 1 sim  \beta  = N $$
$$ N_1 cos  30^0 = \frac {1}{ \sqrt {3} } \times  10 $$
$$ \Rightarrow N_1 = \frac {20}{3} $$
$$ N_1 sin 30^0 = N $$
$$ \Rightarrow N = \frac {20}{3} \times  \frac {1}{2} = \frac {10}{3} N $$

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

From figure

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

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

Let speed of car  = V 
time taken $$ = 42 sec $$
$$  V  \times  42 = 350 $$
$$ V = \frac {350}{42} $$
$$ V = \frac {25}{3}  \times \frac {18}{5} $$
$$ = 30 km / h $$

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 
$$ \frac { 3m(sL)^2}{ 12} + 3m  [ L^2 ] $$
$$ = \frac { 9 mc^2}{4} + 3ml^32 = \frac { 21 ml^2}{4} $$
$$ = \frac {7}{3}  + \frac {21}{4} ml^2 $$

$$ t = 600 \times \frac { 6.5}{ 2} sin \theta  $$
$$ = 600 \times 6.52 \times  \sqrt { 1 -  cos^2 \theta } $$
$$ \tau = 600 \times  \frac { 6.5}{2} \times ( \sqrt { 1 - ( \frac {6}{6.5})^2 } ) = 735 $$
$$ \tau  = 70 N - m  $$
$$ R_2 = mg  = 60 \times 9.8 = 590 $$
$$ R_1 = mR_2 $$
$$ 6.5 R_1 cos  \theta  = 6.09 sin \theta  \times  \frac { 6.5}{ 2} $$
$$R_1 = \frac {1}{26} tan  \theta  = \frac {1}{260} \times  \frac { 2.6}{6} $$
$$ R_1 = \frac {12.59}{6} $$
$$ R_1 = 122.5 N $$

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

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

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

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

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

$$h = u\left(t\right) + at^2$$
$$\implies bt^2 = 10t$$
$$\implies t = 5/3^3$$

$$ x = 5t  $$
$$ V_x = \frac {dx}[dt] = 5  m/s $$
$$  y = 2t^2 + t  $$
$$ V_y = \frac {dy}{dt} =2 \times  2t +1 $$
$$ V_y = 4t + 1 $$
$$ \overrightarrow {V} = \overrightarrow {V_x} \hat {i}   + V_Y  \hat {j} $$
$$ tan  45^0 = \frac {V_y}{V_x} $$
$$ L = \frac {4 t + 1 }{5} $$
$$ 4t +1 = 5 $$
$$ 4t = 4  $$
$$ t = 1 s $$

Formula 
$$ g = \frac { G  \times M}{ R^2} $$
$$ 9.8 = \frac { G \times M}{ R^2} $$
$$ mass J  = 314 M $$
$$ Radius  R  = 11.35 R $$
$$g'= \frac { G \times 314 \times M}{ 11.35 R  \times 11.35 R} $$
$$ = 2.43 \times  \frac {GM}{R^2}  $$
$$ g '= 2.43 \times  9.8 $$
$$ = 23.90 m/s^2 $$

$$\tan\theta = \dfrac{4}{3}$$
$$\theta = \tan^{-1}\left(\dfrac{4}{3}\right)$$
$$\theta = 53^{\circ}$$

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

$$ fact = 20 +10 - 5 $$
$$  =25 N $$
$$ fact - m \times a $$
$$ 25 = 2 \times a $$
$$ a = 12. 5 m/s^2 $$

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)

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

$$m=4kg,H=1m,h=0.8m$$
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

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 
$$ = 175 + 275 $$ 
$$ = 450\ m $$

$$TE = \dfrac{-GMm}{2(R + h)} $$ 

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

$$ = - 1.95 \times 10^{10} J $$ 

$$BE = - TH $$ 
$$ = 1.93 \times 10^{10} J $$

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

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

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

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

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.

$$SO_2$$ is a gas at room temperature.

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.

Force on a spring is given by

$$F= - k \Delta x$$

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

$$24.248 ms^{-1}$$
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$$24.248 ms^{-1}$$

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.

Molar mass of Au = 197 gm/mol

$$10\,N$$

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.

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.

Answer
Positive.
Explanation:
Protons have a positive charge of 1e, which is equal to about $$ 1.602 . 10 ^{-19} C $$

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.

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).

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

$$v=3 \dfrac{m}{s}$$
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 $$P=m.v$$
$$12=4.v$$
$$v=\dfrac{12}{4}$$
$$v=3 \dfrac{m}{s}$$

In $$100\ g$$ of the unknown compound, the ratio of $$C\ ,\ H$$ and $$O$$ is : 

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

A transverse wave is described by equation :
$$y=y_0 \sin 2 \pi \left( ft - \dfrac{x}{\lambda} \right)$$

$$\cos\left(31^{\circ}\right)$$.$$163 m/s = 139.7 m/s$$
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.

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.

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

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

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'

 $$ M_{M Al} =  26.981538 g mol^{-1} $$
$$ M_{M S}  = 32.065 g mol^{-1} $$
$$ M_{M O}=15.9994 g mol^{_1}$$
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
 $$ M_{M Al_{2}} (SO_{4})_{3}  =  342 g mol^{-1} $$

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.

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.

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.

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

$$Answer $$
$$ F_{pull} = 570 N $$ 
$$Explanation :$$

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

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

$$ hcR = 13.6 e V$$

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

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

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

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

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

The number of electrons per coulomb: 

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

$$\vec{f}_{s} \approx 4.8 \times 10^{3}\ N $$ 
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.

(c) The object will continue moving at the same speed. This is brought out by Newton's first law of motion.

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

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. 

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.

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.