Answer:
The moment of inertia I is
I = 2.205x10^-4 kg/m^2
Explanation:
Given mass m = 0.5 kg
And side lenght = 0.03 m
Moment of inertia I = mass x radius of rotation squared
I = mr^2
In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.
Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m
Therefore,
I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2
Answer:
10
Explanation:
(r) = <10 cos 6t, 10 Sin 6t>
The distance traveled by the object is the magnitude of vector r.
The magnitude of vector r is given by


r = 10
Answer:
(a) Angular velocity will be 125.6 rad/sec
(b) Linear velocity will be 144.44 m /sec
(c) Centripetal acceleration = 1849.3031 g
Explanation:
We have given diameter d = 2.30 m
So radius r = 
(a) Speed is given as 1200 rev/min
We know that angular velocity is given by 
(b) Linear speed is given by 
(c) Centripetal acceleration is given by
We know that 
So 
Answer:
u = - 38.85 m/s^-1
Explanation:
given data:
acceleration = 2.10*10^4 m/s^2
time = 1.85*10^{-3} s
final velocity = 0 m/s
from equation of motion we have following relation
v = u +at
0 = u + 2.10*10^4 *1.85*10^{-3}
0 = u + (21 *1.85)
0 = u + 38.85
u = - 38.85 m/s^-1
negative sign indicate that the ball bounce in opposite directon