M = 40 Kg , g=9.8 m/s² , h = 2 m
PE = m g h
PE = (40) (9.8) (2)
PE = 784 J
KE = PE
½m v² = m g h
½ v² = g h
½ v² = (9.8) (2)
½ v² = 19.6
v² = 19.6×2
v² = 39.2
V = √39.2
V = 6.26 m/s
KE = ½mv²
KE = ½(40) (6.26)²
KE =783.8 J
Explanation:
We will assume that the rim of the wheel is also very thin, like the spokes. The distance <em>s</em><em> </em><em> </em>between the spokes along the rim is

The 20-cm arrow, traveling at 6 m/s, will travel its length in

The fastest speed that the wheel can spin without clipping the arrow is

The angular velocity
of the wheel is given by

In terms of rev/s, we can convert the answer above as follows:

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.
Answer: I Believe The Answer Is A. Insulating The Body
For this question, we use the Coulumb's law to calculate the force on each particles. In this law, force between point charges are said to be proportional to product of each charge and is indirectly proportional to the distance of both charges. We do as follows:
F= kq(1)q(2)/d^2
= (9x10^9).(1.41 x 10^-5 C).(-<span>1.41 x 10^-5 C</span><span>)/.44^2
</span> = 4.067 N
Your list that you call "the following properties" is awfully short. / / / The displayed value tells you the car's speed relative to the pavement. (NOT its velocity.)