This next statement is a big deal. It should be up on a board, surrounded
by flashing red and yellow lights, and hung on the wall of every Science
classroom. Although we never see it in our daily lives, it's fundamental to
the workings of the universe, and it's also Newton's first law of motion:
<em>Without friction, it doesn't take <u>ANY</u> force to keep a moving object
moving. </em><em>Force is only required to <u>change</u> the object's speed, or to
<u>change</u> the direction </em><em>in which it's moving.</em>
The answer to the question is: On a level road, and neglecting any friction,
the engine doesn't have to supply ANY force to keep the car going at the
same speed.
1. Frequency 2. measure from trough to trough
Answer:
Explanation:
Given
mass of sphere 
diameter of sphere 
radius 

friction will provide resisting torque so
where 







(b)time taken to decrease its rotational speed by 



Answer:
The center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.
Explanation:
Let the reference point be the center of the Earth

Where;
Xcm is the distance from center of the Earth =?
Me is the mass of the Earth = 6 × 10²⁴ kg
Xe is the center mass of the Earth = 0
Mm is the mass of the moon = 7 × 10²² kg
Xm is the center mass of the moon = 4 × 10⁸ m

Therefore, the center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.
Answer:
Explanation:
Given that,
Mass of counterweight m= 4kg
Radius of spool cylinder
R = 8cm = 0.08m
Mass of spool
M = 2kg
The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:
Then we have,
τ(net) = R~ × T~
τ(net) = R~•i × mg•j
τ(net) = Rmg• k
τ(net) = 0.08 ×4 × 9.81
τ(net) = 3.139 Nm •k
The magnitude of the net torque is 3.139Nm
b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:
L~ = R~ × m~v + I~ω
L = mRv + MR v
L = (m + M)Rv
L = (4 + 2) × 0.08
L = 0.48 Kg.m
C. τ =dL/dt
mgR = (M + m)R dv/ dt
mgR = (M + m)R • a
a =mg/(m + M)
a =(4 × 9.81)/(4+2)
a = 6.54 m/s