Answer: a = 4 m/s²
Explanation:
a = Δv/t = (30 - 18) / 3 = 4 m/s²
Density =mass/volume 120/200 =0.6 g/cm
Orient the semi-circle arc such that it is symmetric with respect to the y-axis. Now, by symmetry, the electric field in the x-direction cancels to zero. So the only thing of interest is the electric field in the y-direction.
dEy=kp/r^2*sin(a) where k is coulombs constant p is the charge density r is the radius of the arc and a is the angular position of each point on the arc (ranging from 0 to pi. Integrating this renders 2kq/(pi*r^3). Where k is 9*10^9, q is 9.8 uC r is .093 m
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Answer:
Option E is correct.
Time the ball remains in the air before striking the ground is closest to 3.64 s
Explanation:
yբ = yᵢ + uᵧt + gt²/2
yբ = 0
yᵢ = 2 m
uᵧ = u sinθ = 20 sin 60 = 17.32 m/s
g = -9.8 m/s², t = ?
0 = 2 + 17.32t - 4.9t²
4.9t² - 17.32t - 2 = 0
Solving the quadratic equation,
t = 3.647 s or t = -0.1112 s
time is a positive variable, hence, t = 3.647 s. Option E.
It is converted into kinetic energy
