Answer:
<em>63.44 rad/s</em>
<em></em>
Explanation:
mass of bullet = 3.3 g = 0.0033 kg
initial velocity of bullet 
= 250 m/s
final velocity of bullet 
= 140 m/s
loss of kinetic energy of the bullet = 
==> 
= 70.785 J
this energy is given to the stick
The stick has mass = 250 g =0.25 kg
its kinetic energy = 70.785 J
from
KE = 
70.785 = 
566.28 = 
= 23.79 m/s
the stick is 1.5 m long
this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m
The angular speed will be
Ω = v/r = 23.79/0.375 =<em> 63.44 rad/s</em>
The answer is B as all the other options contain quantities not related to describing motion
Answer:
40 N
Explanation:
We first need to calculate the acceleration of the tron ball.
Since acceleration, a = (v - u)/t where u = initial velocity of iron ball = 17m/s, v = final velocity of iron ball = 27m/s and t = time taken for the change in velocity = 5 s.
So, a = (v - u)/t
= (27 m/s - 17 m/s)/5 s
= 10 m/s ÷ 5 s
= 2 m/s²
We know force on iron ball, F = ma where m = mass of iron ball = 20 kg and a = acceleration = 2 m/s²
So, F = ma
= 20 kg × 2 m/s²
= 40 kgm/s²
= 40 N
So, the magnitude of the force on the iron ball is 40 N.
