The planetary temperature energy balance is obtained by radiating back the absorbed radiation energy from outer-space, by the planet and thus acquiring thermal equilibrium.
What is the process of attaining thermal equilibrium by Earth?
The Stefan-Boltzmann law states that the more the temperature a planet has, the more it will radiate out to reach thermal equilibrium.
We know that outer space contains large masses of radiative energy freely distributed in its vast expanse. A small fraction of this energy is absorbed by the Earth through the atmosphere, surface land, clouds etc.
Now, radiative balance is achieved when a planet's surface continuously warms up until it reaches its peak at which point the same amount of absorbed energy can then be radiated back to space. The relative amount of energy radiated back by a planet is dependent upon the size of the planet.
A colder planet relatively absorbs lower amount of radiation energy from space. In some time, as the planet heats up enough, the energy is radiated back to the space attaining thermal equilibrium.
Learn more about Stefan-Boltzmann law here:
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Answer:
ΔV=0.484mV
Explanation:
The potential difference across the end of conductor that obeys Ohms law:
ΔV=IR
Where I is current
R is resistance
The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A
R=(pL)/A
Given data
Length L=3.87 cm =0.0387m
Diameter d=2.11 cm =0.0211 m
Current I=165 A
Resistivity of aluminum p=2.65×10⁻⁸ ohms
So
ΔV=IR
ΔV=0.484mV
Answer: ME= E total - E thermal
Answer:
The answer is A. C and O..
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
