Answer:
Speed of ball A after collision is 3.7 m/s
Speed of ball B after collision is 2 m/s
Direction of ball A after collision is towards positive x axis
Total momentum after collision is m×4·21 kgm/s
Total kinetic energy after collision is m×8·85 J
Explanation:
<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>Let the mass of each ball be m kg
v be the velocity of ball A along positive x axis
v be the velocity of ball A along positive y axis
u be the velocity of ball B along positive y axis
Conservation of momentum along x axis
m×3·7 = m× v
∴ v = 3.7 m/s along positive x axis
Conservation of momentum along y axis
m×2 = m×u + m× v
2 = u + v → equation 1
∴ relative velocity of approach = relative velocity of separation
-2 = v - u → equation 2
By adding both equations 1 and 2 we get
v = 0
∴ u = 2 m/s along positive y axis
Kinetic energy before collision and after collision remains constant because it is an elastic collision
Kinetic energy = (m×2² + m×3·7²)÷2
= 8·85×m J
Total momentum = m×√(2² + 3·7²)
= m× 4·21 kgm/s
The minimum input force she'll need to lift the ball is 35 N.
Explanation:
Mechanical advantage of a single pulley is 1. As, she applies 70 N of force to lift the bowling ball, so the output force(weight of the ball) is also 70 N.
Now, adding another pulley gives a mechanical advantage of 2. We have,
M.A = (Output Force)/(Input Force)
Substituting the values we get,
= 35 N
Input force equals to 35 N needs to be applied.
The average current density in the wire is given by:
where I is the current intensity and A is the cross-sectional area of the wire.
The cross-sectional area of the wire is given by:
where r is the radius of the wire. In this problem, , so the cross-sectional area is
and the average current density is