Answer:
100m
Explanation:
100m
s=ut+1/2at^2
s= unknown, u=0, a=2, t=10
s=0*10+1/2(2)(10)^2
s=1/2(2)(100)
s=1(100)
displacement = 100 meters
Work done by force is given by formula

now here work done against friction is given so force of friction here is

It moved by three corridors with each measures

so total distance will be

now from above formula of work done we will say


so above is the work done to move three corridors
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Answer:
(a)0.531m/s
(b)0.00169
Explanation:
We are given that
Mass of bullet, m=4.67 g=
1 kg =1000 g
Speed of bullet, v=357m/s
Mass of block 1,
Mass of block 2,
Velocity of block 1,
(a)
Let velocity of the second block after the bullet imbeds itself=v2
Using conservation of momentum
Initial momentum=Final momentum







Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s
(b)Initial kinetic energy before collision



Final kinetic energy after collision



Now, he ratio of the total kinetic energy after the collision to that before the collision
=
=0.00169