Answer:
The new resistance comes out to be = 4 times of original resistance .
Answer:
The direction of angular velocity and angular momentum are perpendicular to the plane of rotation. Using the right hand rule, the direction of both angular velocity and angular momentum is defined as the direction in which the thumb of your right hand points when you curl your fingers in the direction of rotation.
Explanation:
Answer:
0.1308
Explanation:
To keep the rider from sliding down, then the friction force
must at least be equal to gravity force 


where μ is the coefficient, N is the normal force acted by the rotating cylinder, m is the mass of a person and g = 9.81 m/s2 is the gravitational acceleration.
According to Newton's 3rd and 2nd laws, the normal force would be equal to the centripetal force
, which is the product of centripetal acceleration
and object mass m

Therefore


The centripetal acceleration is the ratio of velocity squared and the radius of rotation

Therefore

The toy rocket is launched vertically from ground level, at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance.
The total energy of the rocket, which is a sum of its kinetic energy and potential energy, is constant.
At a height of 72 m with the rocket moving at 30 m/s, the total energy is m*9.8*72 + (1/2)*m*30^2 where m is the mass of the rocket.
At ground level, the total energy is 0*m*9.8 + (1/2)*m*v^2.
Equating the two gives: m*9.8*72 + (1/2)*m*30^2 = 0*m*9.8 + (1/2)*m*v^2
=> 9.8*72 + (1/2)*30^2 = (1/2)*v^2
=> v^2 = 11556/5
=> v = 48.07
<span>The velocity of the rocket when it impacts the ground is 48.07 m/s</span>
final velocity = 0
acceleration = - 10 m/ s 2
distance. = 20 m
u = ?
v^2 - u ^2 = 2 a s
0^2 - u^ 2 = 2 * -10 * 20
-u^2 = -400
u = √ 400
u = 20m / s