Answer:
The correct option is;
Force of Friction
Explanation:
As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path
The motion equation is
v = ωr and we have the centripetal acceleration given by
α = ω²r and therefore centripetal force is then
m×α = m × ω²r = m × v²/r
The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.
Answer
Time period T = 1.50 s
time t = 40 s
r = 6.2 m
a)
Angular speed ω = 2π/T
=
= 4.189 rad/s
Angular acceleration α = 
= 
= 0.105 rad/s²
Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²
b)The maximum speed.
v = 2πr/T
= 
= 25.97 m/s
So centripetal acceleration.
a = 
= 
= 108.781 m/s^2
= 11.1 g
in combination with the gravitation acceleration.


<span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365
Fsf = Static frictional force = (coefficient of static friction) * (Normal force)
So the least for you could exert to move it is equal to the Fsf.
Fsf = (0.49)(1365N)</span><span>
</span>

Explanation:
Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as
(1)
Assuming that the velocity remains constant then

Solving for
we get

Before we plug in the given values, we need to convert them first to their appropriate units:
The thrust <em>F</em><em> </em> is

The exhaust rate dm/dt is


Therefore, the velocity at which the exhaust gases exit the engines is


Answer:
Ans is 200 J
Explanation:
Given: Force = 20N
Distance = 10m
Work done = Force * displacement
= 20 * 10
= 200 J