The solution for this problem is:
Let u denote speed.
Equating momentum before and after collision:
= 0.060 * 40 = (1.5 + 0.060) u
= 2.4 = 1.56 u
= 2.4 / 1.56 = 1.56 u / 1.56
= 1.6 m / s is the answer for this question. This is the speed after the collision.
Explanation:
It is given that,
Mass of the rim of wheel, m₁ = 7 kg
Mass of one spoke, m₂ = 1.2 kg
Diameter of the wagon, d = 0.5 m
Radius of the wagon, r = 0.25 m
Let I is the the moment of inertia of the wagon wheel for rotation about its axis.
We know that the moment of inertia of the ring is given by :
The moment of inertia of the rod about one end is given by :
l = r
For 6 spokes, 
So, the net moment of inertia of the wagon is :
So, the moment of inertia of the wagon wheel for rotation about its axis is 
. Hence, this is the required solution.
They require a medium to travel through
Answer:
t = 1.16 s.
Explanation:
Given,
speed of conveyor belt, v = 3.2 m/s
coefficient of friction,f = 0.28
Using newton second law
f = ma
and we also know that frictional force
f = μ N
f = μ m g
equating both the force equation
a = μ g
a = 0.28 x 9.81
a = 2.75 m/s²
Using Kinematic equation
v = u + at
3.2 = 0 + 2.75 x t
t = 1.16 s.
Time taken by the box to move without slipping is 1.16 s.
Answer:
-1.5m/s²
Explanation:
Acceleration can be thought of as [Change in Velocity]/[Change in time]. To find these changes, you simply subtract the initial quantity from the final quantity.
So for this question you have:
- V_i = 110m/s
- V_f = 80m/s
- t_i = 0s
- t_f = 20s
which means that the acceleration = (80-110)/(20-0)[m/s²] = (-30/20)m/s² = -1.5m/s²
