Answer:
= -32.53 m / s
this velocity is directed downwards
Explanation:
This is a free fall exercise, let's use the expression
= v_{oy}^{2} + 2 g (y -yo)
where we are assuming that there is friction with the air, as the body falls its initial velocity is zero
v_{oy} = √ 2g (y - y₀)
let's calculate
v_{y} = √ (2 9.8 (0-54.0))
= -32.53 m / s
this velocity is directed downwards
It's a form of mechanical energy
Answer:
B
Explanation:
Gravitational Energy is the energy of position or place. A rock resting at the top of a hill contains gravitational Potential energy. Hydropower, such as water in a reservoir behind a dam, is an example of gravitational potential energy.
By using drift velocity of the electron, the current flow is 7.20 ampere.
We need to know about drift velocity of electrons to solve this problem. The drift velocity can be determined as
v = I / (n . A . q)
where v is drift velocity, I is current, n is atom number density, A is surface area and q is the charge.
From the question above, we know that
d = 2.097 mm
r = (0.002097 / 2) m
v = 1.54 mm/s = 0.00154 m/s
ρ = 8.92 x 10³ kg/m³
q = e = 1.6 x 10¯¹⁹C
Find the atom density
n = Na x ρ / Mr
where Na is Avogadro's number (6.022 x 10²³), Mr is the atomic weight of copper (63.5 g/mol = 0.635 kg/mol).
n = 6.022 x 10²³ x 8.92 x 10³ / 0.635
n = 8.46 x 10²⁷ /m³
Find the current flows
v = I / (n . A . q)
0.00154 = I / (8.46 x 10²⁷ . πr² . 1.6 x 10¯¹⁹)
0.00154 = I / (8.46 x 10²⁷ . π(0.002097 / 2)² . 1.6 x 10¯¹⁹)
I = 7.20 ampere
For more on drift velocity at: brainly.com/question/25700682
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Your being on the moon has no effect on the moon's
gravitational field strength, or on the Earth's for that
matter.
However, YOU notice a change on YOU when YOU move
from one to the other, because of the effect of the gravitational
field strength on you and your internal organs.
If you could stand on the moon, you would experience an incredible
sense of lightness, since the forces of attraction between the moon
and anything else are only 16% as great as the same forces are on
Earth.
