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HACTEHA [7]
3 years ago
13

How does the earth release energy back into the atmosphere?​

Physics
2 answers:
LuckyWell [14K]3 years ago
5 0

Answer:

by releasing it lol? it does it from like the thing and thing and yeah thing

Allushta [10]3 years ago
3 0

Answer:

Explanation:

The earth-atmosphere energy balance is the balance between incoming energy from the Sun and outgoing energy from the Earth. Energy released from the Sun is emitted as shortwave light and ultraviolet energy. When it reaches the Earth, some is reflected back to space by clouds, some is absorbed by the atmosphere, and some is absorbed at the Earth's surface.

Hope this helped you!

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Find the change in thermal energy of a 25kg severed clown doll head that heats up from 25°C to 35°C, and has the specific heat o
timama [110]

Answer:

Q = 425 kJ

Explanation:

Given that,

Mass, m = 25 kg

The clown doll head that heats up from 25°C to 35°C

The specific heat is 1700 J/kg°C

We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :

Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ

So, 425 kJ of thermal energy is severed.

6 0
3 years ago
The kinetic energy of an object can sometimes be greater than a potential energy a originally possessed, true or false?
Mamont248 [21]

Answer:

true

Explanation:

Energy stored in the nuclei of atoms can be used to generate electricity. ... Most of the energy of food is converted to heat.

4 0
3 years ago
Read 2 more answers
Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half
lutik1710 [3]
The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
v= \frac{S}{t}

The total distance covered is
S=200 km+0+100 km=300 km
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
t=2h+0.5h+1.5 h=4 h
Therefore, the average velocity is 
v= \frac{S}{t}= \frac{300 km}{4 h}=75 km/h
0 0
3 years ago
An airplane has wings, each with area A, designed so that air flows over the top of the wing at 265 m/s and underneath the wing
exis [7]

Answer

given,

Pressure on the top wing = 265 m/s

speed of underneath wings = 234 m/s

mass of the airplane =  7.2 × 10³ kg

density of air =  1.29 kg/m³

using Bernoulli's equation

 P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2

 \Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)

 \Delta P =\dfrac{1}{2}\times 1.29\times (265^2-234^2)

 \Delta P =9977.5 Pa

Applying newtons second law

2 Δ P x A - mg = 0

A =\dfrac{mg}{2\Delta P}

A =\dfrac{7.2\times 10^3 \times 9.8}{2\times 9977.5}

    A = 3.53 m²

7 0
3 years ago
Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o
Mila [183]

Answer:

a) x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) x = 4.47 cm

c) x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm

c) The center of the mass of the beads realtive to the center of bead is:

x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

d) x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm

6 0
3 years ago
Read 2 more answers
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