Since
Electric potential energy = qV
Where V = Ed
Hence
Electric potential energy = q(Ed) --- (1)
Since E = 1.0 * 10^3 N/C
d = 0.10 m
q = 4 * 10^-6 C
Plug in the values in (1)
(1) => Electric potential energy = 4 * 10^-6(1.0 * 10^3 * 0.10)
Electric potential energy = 400 μJ
Answer:
Part a)
Part b)
Direction = upwards
Explanation:
When ball is dropped from height h = 4.0 m
then the speed of the ball just before it will strike the ground is given as
Now ball will rebound to height h = 2.00 m
so the velocity of ball just after it will rebound is given as
Part a)
Average acceleration is given as
Part B)
As we know that ball rebounds upwards after collision while before collision it is moving downwards
So the direction of the acceleration is vertically upwards
Answer:
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Explanation:
Answer:
V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s
1/2 m V^2 = m g h conservation of energy
h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m max height
Check:
t = 28.2 / 9.8 = 2.88 sec time to reach max height
h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m
Answer:b
Explanation: If you look at the line on the graph, you can see that it is going downward, meaning it has a negative slope, and choice b is the only one that has a negative slope