Answer:
I will answer in English.
Here we will use the relation
Velocity*time = distance
So:
a) velocity = 3m/s
time = 2s
Distance = 3m/s*2s = 6m
b) velocity = 2m/s
time = 3.5s
Distance = 2m/s*3.5s = 7m
c) velocity = 10m/s
time = 0.5s
Distance = 10m/s*0.5s = 5m
d) velocity = 4m/s
time = 2.5s
Distance = 4m/s*2.5s = 9m
e) velocity = 1.5m/s
time = 5s
Distance = 1.5m/s*5s = 7.5m
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:
179.655m
Explanation:
Given
Maximum speed of the arrow v = 60m/s
Time taken to hit the top of the cliff t = 7.0s
Required
Height of the cliff H
Using the equation of motion
H = vt + 1/2gt²
Substitute into the formula:
H = 60(7) + 1/2 (-9.81)(7²) (g is negative due to upward motion of the arrow)
H = 420-4.905(49)
H = 420-240.345
H = 179.655m
Hence the cliff is 179.655m high
In space there is a vacuum, light travels as photons, and the see a photon it needs to bounce off of a surface to make it shine, so since the earth has an atmosphere, the light shines.
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