1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.
2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.
3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.
\Delta L= \alpha L_0 (T_f-T_i)
= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
So your answer is B.
Answer:
(a) 62.5 m
(b) 7.14 s
Explanation:
initial speed, u = 35 m/s
g = 9.8 m/s^2
(a) Let the rocket raises upto height h and at maximum height the speed is zero.
Use third equation of motion


h = 62.5 m
Thus, the rocket goes upto a height of 62.5 m.
(b) Let the rocket takes time t to reach to maximum height.
By use of first equation of motion
v = u + at
0 = 35 - 9.8 t
t = 3.57 s
The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.