A 2.0kg object is dropped from a height of 30m. After it drops for 2.0 seconds, what is its kinetic
energy and what is its potential energy?
(Assume no air resistance.)

1) The kinetic energy of the object after it drops for 2.0 seconds is approximately 384.9 Joules

2) The potential energy of the object after it drops for 2.0 seconds is approximately 204 J

Explanation:

1) The given mass of the object, m = 2.0 kg

The height from which the object is dropped, h = 30 m

The kinetic energy of the object after it drops for 2.0 seconds = Required

Kinetic energy, K.E. = (1/2)·m·v²

Where;

v = The velocity of the object

The kinematic equation for finding the velocity of the object is presented as follows;

v = u + g·t

Where;

u = The initial velocity of the object = 0

g = The acceleration due to gravity of the object ≈ 9.81 m/s²

t = The time of motion of the object = 2.0 seconds

∴ The velocity after 2 seconds, v ≈ 0 + 9.81 m/s² × 2 s = 19.62 m/s

The kinetic energy, K.E. after 2 seconds as the object drops is given as follows;

= (1/2) × 2.0 kg × (19.62 m/s)² = 384.9444 J ≈ 384.9 J

2) The total energy, M.E. of the object at the top, h = 30 m, u = 0, is given as follows;

The total mechanical energy, M.E. = P.E. + K.E.

M.E. = m·g·h + (1/2)·m·u²

∴ M.E. = 2.0 kg × 9.81 m/s² × 30 m + 0 = 588.6 J

M.E. = 588.6 J

Given that the total mechanical energy, M.E., is constant, we have;

At 2.0 seconds, M.E. = 588.6 J , K.E. ≈ 384.9 J, P.E. = M.E. - K.E.

∴ P.E. = 588.9 J - 384.9 J ≈ 204 J

The potential energy after it drops 2.0 seconds, P.E. ≈ 204 J

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.