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Rama09 [41]
3 years ago
12

A horse runs for 15 seconds to the south at a speed of 12 m/s. What is the

Physics
1 answer:
elixir [45]3 years ago
8 0

Answer: 180 m south

Explanation:

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Light of different colors is emitted from different stars because :.
Varvara68 [4.7K]

Answer:

Different temperatures

Explanation:

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2 years ago
The _______ is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. A.
Vladimir [108]

"A is correct answer." The effective length of the tube is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. "Hope this helps!" "Have a great day!" "Thank you for posting your question!"

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3 years ago
At which moon position would a person on Earth see the entire half of the moon(full-moon)?
jeka94
Sun-earth-moon in a straight line. Earth in the 'middle'.
3 0
3 years ago
Read 2 more answers
Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.
Tcecarenko [31]
The energy of a photon is given by
E=hf
where
h=6.6 \cdot 10^{-34} Js is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is 
f=5.49 \cdot 10^{14}Hz
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J
8 0
3 years ago
A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling
KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

      \frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}} ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

     \frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} ...... (2)

Now, equating both equations (1) and (2) as follows.

 Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} = Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}        

        \gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}

                    = \frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})

                    = \frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})

                    = 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

4 0
3 years ago
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