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storchak [24]
3 years ago
12

Part of the Sun's energy that reaches the Earth's surface is absorbed by land and water as heat. The Earth's surface then releas

es heat back to the atmosphere. Which of the following is true about this solar energy that is absorbed and released by Earth's surface as heat?
All of this heat escapes the Earth's atmosphere, which keeps the planet hot.

Much of the heat is trapped low in the atmosphere and is a major factor in determining Earth's climate.

Much of the heat is trapped at the top of Earth's atmosphere to protect Earth from ultraviolet radiation.

All of this heat escapes the Earth's atmosphere, which keeps the planet c
Physics
2 answers:
DochEvi [55]3 years ago
5 0

Answer: B. Much of the heat is trapped low in the atmosphere and is a major factor in determining Earth's climate.

Explanation:

Presence of atmosphere on Earth is one of the major reason of life being habitable on it. Earth receives large amount of solar radiation from the Sun. 30% of this reflects back and rest is absorbed by clouds, oceans and land mass. The heat trapped by the atmosphere makes the conditions livable on it. An average temperature of 14 degrees is maintained with the heat trapped. In the low atmosphere, this trapped heat is a major factor determining climate in different hemispheres.

Thus, correct option is B. Much of the heat is trapped low in the atmosphere and is a major factor in determining Earth's climate.

Feliz [49]3 years ago
4 0
I think the correct answer is B- heat is trapped in the earth atmosphere 
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An airplane is flying north. What is the direction of the air friction force acting or
Nimfa-mama [501]

Answer: south

Explanation:

8 0
3 years ago
The pin symbolizing the greenhand ffa degree is made of what type of metal and why
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5 0
4 years ago
A projectile is launched at ground level with an initial speed of 50.0m/s at an angle of 30° above the horizontal. It strikes a
tino4ka555 [31]

Answer:

x = 129.9 m

y = 30.9 m

Explanation:

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

v_{i} = 50 m/s

Angle = 30°

Time = t = 3 s

horizontal component of velocity = v_{i_{x}} = v_{i}cos30°

v_{i_{x}} = 50cos30°

v_{i_{x}} = 43.3 m/s

Vertical component of velocity = v_{i_{y}} = v_{i}Sin30°

v_{i_{y}} = 50Sin30°

v_{i_{y}} = 25 m/s

This is a projectile motion, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

But the vertical component of velocity varies with time and there is an acceleration along vertical direction which is equal to gravitational acceleration g.

Horizontal distance = x =  v_{i_{x}}t

x =  43.3*3

x = 129.9 m

Vertical Distance = y = v_{i_{y}}t -0.5gt²

y = 25*3 - 0.5*9.8*3²

y = 75 - 44.1

y = 30.9 m

3 0
3 years ago
An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with
Ket [755]

Answer:

Part a)

I = 1879.7 kg m^2

Part b)

\alpha = 0.70 rad/s^2

Part c)

I = 153.8 kg m^2

Part 4)

angular acceleration will be ZERO

Part 5)

I = 345.6 kg m^2

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)

So we have

I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)

I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)

I = (\frac{16}{3} + 127)m_r R^2

I = (\frac{16}{3} + 127)(6.85)(1.44)^2

I = 1879.7 kg m^2

Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

\tau = F\frac{L}{2}

\tau = 460(2R)

\tau = 460 \times 2 \times 1.44

\tau = 1324.8 Nm

now angular acceleration is given as

\alpha = \frac{\tau}{I}

\alpha = \frac{1324.8}{1879.7}

\alpha = 0.70 rad/s^2

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)

x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}

so moment of inertia about this position is given as

I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)

so we have

I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})

I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})

I = 6.85(1.44)^2\times 10.83

I = 153.8 kg m^2

Part 4)

If force is applied parallel to the length of rod

then we have

\tau = \vec r \times \vec F

\tau = 0

so angular acceleration will be ZERO

Part 5)

moment of inertia about right edge of the sphere is given as

I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)

so we have

I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)

I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)

I = 6.85(1.44)^2\times 24.33

I = 345.6 kg m^2

6 0
3 years ago
A train departs from its station at a constant acceleration of 5m/s2. What is the speed of the train at the end of 20 seconds
Shalnov [3]

a = v/t

v = a.t

v = 5 * 20 = 100m/s

8 0
4 years ago
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