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german
2 years ago
13

Heat transfers through Earth's systems in different ways. In which atmospheric action can we see evidence of conduction? Cold ai

r pushes warm air upward, creating a current. Cold air pushes warm air upward, creating a current. The surface of the Earth heats the air that contacts it. The surface of the Earth heats the air that contacts it. Radiation from the Sun heats the surface of the Earth. Radiation from the Sun heats the surface of the Earth. Air increases in density and sinks back towards the Earth.
Physics
1 answer:
AlladinOne [14]2 years ago
8 0

Answer:

B. The surface of the Earth heats the air that contacts it.

Explanation:

Conduction is a mode of heat transfer in which energy is transferred from a hot molecule to the neighboring molecules due to contact. Thus, heat energy is transferred from one molecule to another in the process.

Therefore, the air that comes in contact with the surface of the earth heats up by conduction. Since the surface of the earth is heated up due to the radiation from the Sun.

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A child sleds down a hill with an acceleration of 2.94 m/s2. If her initial speed is 0.0 m/s and her final speed is 17.5 m/s, ho
hram777 [196]

Answer:

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

Explanation:

Given that the child accelerates uniformly and that both initial (v_{o}) and final speeds (v_{f}), measured in meters per second, and acceleration (a), measured in meters per square second, are known, we proceed to use the following kinematic equation to determine the time taken to travel from the top of the hill to the bottom (t), measured in seconds, is:

t = \frac{v_{f}-v_{o}}{a} (1)

If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 17.5\,\frac{m}{s} and a = 2.94\,\frac{m}{s^{2}}, then the time taken is:

t = \frac{17.5\,\frac{m}{s}-0\,\frac{m}{s} }{2.94\,\frac{m}{s^{2}} }

t = 5.952\,s

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

8 0
3 years ago
A figure shows a vertically moving block on the end of a cord. The graph next to the figure gives the vertical velocity componen
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3 years ago
A certain radio wave has a wavelength of 6.0 × 10-2m. What is its frequency in hertz?
rodikova [14]

Answer:

The frequency of the wave is 5 x 10⁹ Hz

Explanation:

Given;

wavelength of the radio wave, λ = 6.0 × 10⁻²m

radio wave is an example of electromagnetic wave, and electromagnetic waves travel with speed of light, which is equal to 3 x 10⁸ m/s².

Applying wave equation;

V = F λ

where;

V is the speed of the wave

F is the frequency of the wave

λ  is the wavelength

Make F the subject of the formula

F = V /  λ

F = (3 x 10⁸) / (6.0 × 10⁻²)

F = 5 x 10⁹ Hz

Therefore, the frequency of the wave is 5 x 10⁹ Hz

8 0
3 years ago
A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.
34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated (Q), measured in joules, by the cylindrical resistor is the power multiplied by operation time (\Delta t), measured in hours. That is:

Q = \dot Q \cdot \Delta t (1)

If we know that \dot Q = 1.2\,W and \Delta t = 86400\,s, then the amount of heat dissipated by the resistor is:

Q = (1.2\,W)\cdot (86400\,s)

Q = 103680\,J

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux (Q'), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (A), measured in square meters:

Q' = \frac{\dot Q}{A} (2)

Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h } (3)

Where:

D - Diameter, measured in meters.

h - Length, measured in meters.

If we know that \dot Q = 1.2\,W, D = 4\times 10^{-3}\,m and h = 2\times 10^{-2}\,m, the heat flux of the resistor is:

Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }

Q' \approx 4340.589\,\frac{W}{m^{2}}

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (r), no unit, is the ratio of the top and bottom surfaces to total surface:

r = \frac{\frac{\pi}{2}\cdot D^{2}}{A} (3)

If we know that A \approx 2.765\times 10^{-4}\,m^{2} and D = 4\times 10^{-3}\,m, then the fraction is:

r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}

r = 0.045

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0
3 years ago
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