Answer:
3k mph
Explanation:
don't take my answer it is wrong
1) Upward motion
y1 = Vo*t - g(t^2) / 2
2) Downward motion
y2 = D - [Vo*t + g(t^2) / 2]
3) Collision => y1 = y2
y1 = Vo*t - g(t^2) / 2
y2 = D - [Vo*t + g(t^2) / 2]
Vo*t - g(t^2) / 2 = D - [Vo*t + g(t^2) / 2]
Vo*t - g(t^2) / 2 = D - Vo*t - g(t^2)/2
D = 2Vo*t => t = D / (2Vo)
Substitute the value of t in the equation of y1 (it is the same if you do it in the equation of y2)
y1 = Vo*t - g(t^2) / 2 = Vo [D/(2Vo) ] - g [D / (2Vo)]^2 / 2
y1 = D/2 - g(D^2) / 8(Vo ^2)
Answer: y1 = D/2 - g(D^2) / 8(Vo ^2)
R = ρl/A
From the equation above R = Resistance, l = length, A = Cross Sectional Area of wire.
From the equation, it can be seen that R would increase if the wire's area is reduced.
If the area of the wire is reduced, means the same thing as:
<span>A. decreasing the wire’s thickness</span>
To what? You did not put the question.