Explanation
The average speed is given as,
vavg=st
vavg=V22α+V22βVα+Vβ
vavg=V2
When a body moves upward the acceleration due to gravity is,
g′=g+a
Hence, the pressure is
P=h0ρ(g+a)
=h0(g+a)g
Hence, the barometer reading will be more than the measured value at rest.
The momentum of each bullet is given as,
Pb=mbvb
=10×10−3×500
=5kgm/s
The force on the vehicle will be equal to change in momentum in 1sec.
The force is given as,
F=nPb
=10×5
=50N
The acceleration of the vehicle is given as,
a=Fmv
=502000
=0.025m/s2
Thus, the acceleration produced in the vehicle is 0.025m/s2.
When the system is released from the position of rest, the equations of motions are given as,
4a=4g−T (1)
T=3a (2)
From the equation (1) and (2) it can be written as,
7a=4g
a=4g7
a≈g2
Thus, the acceleration of block B is g2 towards the left side.
Let,
Constant force =K1
Variable force =−K2V
Net Force F=K1−K2V
Acceleration a=K1−K2Vm
As the velocity increases the net force will decrease. When both retarding and constant forces become equal then net force is zero. Hence the velocity is maximum and acceleration will increase from zero to constant.
The acceleration is given as
v=u+at
a=2m/s2
The force applied on the horizontal surface is such that it just overcome the static friction
F−μN=ma
200−μ⋅30⋅10=30×2
The static friction is0.47
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: I tonne 1000 kg.)
Displacement in first, 10sec
s1=12a1t12=12×2×102=100m
Speed achieve in first 10sec
v1=u1+a1t1=0+2×10=20m/s
Displacement in 30sec, acceleration is zero.
s2=v1t2=20×30=600m
Displacement in last due to deacceleration, a3=−4m/s2
v2−v12=2as
s3=0−2022×(−4)=50m
Net displacement
Stotal=s1+s2+s3=100+600+50=750m
Total displacement is 750m
The time taken to reach the stone on the floor is given as,
s=ut+12(a+g)t2
3=8×t+12(2+10)t2
3=8×t+6t2
t=0.3052sec
Since 50N force is required for the car to move with constant velocity.
Additional force applied to move a car of 500kg with acceleration of 1m/s2is given by
F=ma
F=500×1
F=500N
Thus total force applied
FT=500+50
F=550N
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