Gravitational potential energy, relative to some level =
(mass of the object)
times
(height above the reference level)
times
(acceleration due to gravity) .
Explanation:
Given:
v₀ = 0 m/s
a = 3 m/s²
t = 4 s
Find: Δx and v
Δx = v₀ t + ½ at²
Δx = (0 m/s) (4 s) + ½ (3 m/s²) (4 s)²
Δx = 24 m
v = at + v₀
v = (3 m/s²) (4 s) + 0 m/s
v = 12 m/s
True, an object at rest stays and rest and an object in motion stays in motion
<span>Thermocline is a layer between
warm water from the ocean’s surface and cool water from below the ocean. In here,
the temperature decreases rapidly from the warmer layer to the colder layer. A thermocline forms due to the heat of the sun
heating the ocean’s surface. Because of the difference in density between warm
and cooler ocean water, cooler ocean water sinks and warmer ocean water floats.
This is caused due to the heat and mass transfer between particles of the
ocean. The answer is letter C. The sun’s radiation does not extend below a
certain depth; therefore, deeper ocean water is colder than surface water.</span>
Answer:
ΔU = 5.21 × 10^(10) J
Explanation:
We are given;
Mass of object; m = 1040 kg
To solve this, we will use the formula for potential energy which is;
U = -GMm/r
But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.
Thus;
ΔU = -GMm((1/r_f) - (1/r_i))
Where;
M is mass of earth = 5.98 × 10^(24) kg
r_f is final radius
r_i is initial radius
G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²
Since, it's moving to altitude four times the Earth's radius, it means that;
r_i = R_e
r_f = R_e + 4R_e = 5R_e
Where R_e is radius of earth = 6371 × 10³ m
Thus;
ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)
× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))
ΔU = 5.21 × 10^(10) J
