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:
<h2>66.67 km/hr</h2>
Explanation:
The average velocity of the car can be found by using the formula

d is the distance
t is the time taken
From the question we have

We have the final answer as
<h3>66.67 km/hr</h3>
Hope this helps you
From the graph, it can be seen that the constant force that John exerted in order to move the object is 14N. Work is calculated by multiplying the force with the distance to which the object moves in parallel with the direction of the force.
Work = Force x displacement
Work = (14 N) x (8 m)
Work = 112 J
The closest value is 110J. Thus, the answer to this item is the second choice.
Answer:
4.9 x 10^-19 J, 2.7 x 10^-19 J
Explanation:
first wavelength, λ1 = 410 nm = 410 x 10^-9 m
Second wavelength, λ2 = 750 nm = 750 x 10^-9 m
The relation between the energy and the wavelength is given by
E = h c / λ
Where, h is the Plank's constant and c be the velocity of light.
h = 6.63 x 10^-34 Js
c = 3 x 10^8 m/s
So, energy correspond to first wavelength
E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J
E1 = 4.9 x 10^-19 J
So, energy correspond to second wavelength
E2 = (6.63 x 10^-34 x 3 x 10^8) / (750 x 10^-9) = 2.652 x 10^-19 J
E2 = 2.7 x 10^-19 J