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2 years ago

12

What happens to solar radiation when it is absorbed

1 answer:

Zanzabum2 years ago

8 0

Answer:

Absorbed sunlight is balanced by heat radiated from Earth's surface and atmosphere. ... The atmosphere radiates heat equivalent to 59 percent of incoming sunlight; the surface radiates only 12 percent. In other words, most solar heating happens at the surface, while most radiative cooling happens in the atmosphere

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Which of the following is the correct SI unit to measure of rope?

sergij07 [2.7K]

Hello!

The answer should be the third option.

Explanation: Because the meters is the correct SI unit to measure of rope. And thank you for posting at here on Brainly. And have a great day! -Charlie

7 0

2 years ago

The kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s.

eimsori [14]

E=(mV^2)/2
m=1000kg, V=20m/s
then, E=(1000kg*(20m/s)^2)/2
E=(1000*400)/2 J = 200000J

8 0

3 years ago

Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statemen

muminat

Answer:

Option A

Explanation:

From the question we are told that:

Mass $m=0.20kg$

Velocity $v=4m/s$

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

$M_{a1}=mV$

$M_{a1}=0.2*4$

$M_{a1}=0.8$

Final Momentum

$M_{a2}=-0.8kgm/s$

Therefore

$\triangle M_a=-1.6kgm/s$

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

$M_{b1}=mV$

$M_{b1}=0.2*4$

$M_{b1}=0.8$

Final Momentum

$M_{b2}=-0 kgm/s$

Therefore

$|\triangle M_a|>|\triangle Mb|$

Option A

4 0

3 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

4 0

2 years ago

The PE of the box that is on a 2.0 m high self is 1600 J. What is the power expelled to lift the box to this height in 10.0 seco

Pani-rosa [81]

Answer:

160 W

Explanation:

Power is the ratio of work to time:

(1600 J)/(10 s) = 160 J/s = 160 W

7 0

2 years ago