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Otrada [13]
3 years ago
15

What is the earths energy budget? A. The balance between the amount of energy coming into earth from the sun and going back out

into space B. The total energy available in the universe minus earths energy use C. The total energy output by the sun minus all the solar system plants energy use D. The total energy output by the sun plus the energy reflected by earths atmosphere back into space
Physics
2 answers:
OlgaM077 [116]3 years ago
7 0
Our Earth budget is how much energy we receive from the sun.
So the best answer is A
myrzilka [38]3 years ago
5 0
<h3 />

A. The balance between the amount of energy coming into earth from the sun and going back out into space

<h3 />
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What is the difference between a series circuit and a parallel circuit?
Angelina_Jolie [31]
Parallel has more than one circuit or form of energy

series has only one form of energy circuit
6 0
3 years ago
(8c7p26) During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with lar
likoan [24]

Answer: 1175 J

Explanation:

Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."

Given

Spring constant, k = 102 N/m

Extension of the hose, x = 4.8 m

from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m

Work done =

W = 1/2 k [x(i)² - x(f)²]

Since x(f) = 0, then

W = 1/2 k x(i)²

W = 1/2 * 102 * 4.8²

W = 1/2 * 102 * 23.04

W = 1/2 * 2350.08

W = 1175.04

W = 1175 J

Therefore, the hose does a work of exactly 1175 J on the balloon

7 0
3 years ago
the carnot cycle attempts to model the most efficient possible process by avoiding what? adiabatic processes isothermal processe
kobusy [5.1K]

The carnot cycle attempts to model the most efficient possible process by avoiding irreversible processes.

In essence, the Carnot cycle is a reversible cycle made up of four other reversible processes. A reversible process is one that can be thought of as consisting of a sequence of equilibrium stages because it is carried out endlessly slowly.

Essentially, this means that any reversible cycle can be performed in reverse and that the amount of work or heat exchanged along the forward and backward pathways is the same.

It goes without saying that such reversible processes are not possible because they would take an unlimited amount of time. Therefore, the Carnot Engine is described as an idealized heat engine that uses the Carnot Cycle, a reversible cycle.

Learn more about carnot cycle here;

brainly.com/question/13040188

#SPJ4

5 0
9 months ago
A person walking covers 5 m ikn 10 s how fast is the person moving
aniked [119]

Not what I'd call 'fast' at all.

Speed = (distance covered) / (time to cover the distance) .

Speed = (5 meters) / (10 seconds)

<em>Speed = 0.5 meter per second</em> .

That's like about 1.1 mile per hour .

Normal walking speed is considered to be around 1.4 m/s ... about 3.1 mph, or 14 meters in 10 seconds.

I've got a grandson who hasn't even turned 1 yet.  He crawls and  doesn't walk, but if you only cover 5m in 10s, he'd leave you in the dust pretty quick.

5 0
3 years ago
The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a
s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = 0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg

r = Radius of orbit of planet = 0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m

v = Orbital speed

The kinetic and potential energy balance is given by

\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days  

The orbital period is 14.45 days

5 0
2 years ago
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