Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
The angle of inclination is calculated using sin
function,
sin θ = 5 m / 20 m = 0.25
θ = 14.4775°
<span>The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25 </span>
F net = 49N
<span>Work is product of net force and distance:
W = F net * d = 49 * 20 </span>
<span>Work = 980 J </span>
Answer:
yes
Explanation:
I think so because it not mention in the law
Answer:
s
Explanation:
From the question we are told that
The outer ring with a radius of 30 m
inner Gravity Approximately 9.80 m/s'
Outer Gravity Approximately 5.35 m/s.
Generally the equation for centripetal force is given mathematically as
Centripetal acceleration enables Rotation therefore?
Considering the outer ring,
Therefore solving for Period T
Generally the equation for solving Period T is mathematically given as
s
