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ddd [48]
2 years ago
11

The acceleration due to gravity on the moon is 1.6 m/s2, about a sixth that of Earth’s. Which accurately describes the weight of

an object on the moon? An object on the moon is 1/6 times lighter than on Earth. An object on the moon is 1/6 times heavier than on Earth. An object on the moon is six times lighter than on Earth. An object on the moon is six times heavier than on Earth.
Physics
2 answers:
Aleks04 [339]2 years ago
7 0

it is C i just took it on the test online

tatyana61 [14]2 years ago
6 0

The answer is C.  

An object on the moon is six times lighter than on Earth.


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A reaction that releases heat is:<br> endothermic<br> exothermic<br> none of the above
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Answer:

Exothermic reaction

Explanation:

  • A reaction that releases heat is known as an exothermic reaction.
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  • A reaction that absorbs or gains heat from the surrounding is known as endothermic reaction.
  • On the other hand, if a reaction releases heat to the surrounding is known as exothermic reaction.
  • An exothermic reactions results to an increase in temperature of a solution while an endothermic reaction results to a decrease in the temperature of a solution.
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Describe an object's velocity when an acceleration-time graph is zero?
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Anything times zero is zero
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How do gravitational and electic force compare
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Electric Forces. ... Just like objects that have mass exert gravitational forces on each other, objects that are charged will also exert electric forces on each other. The electric force is directly proportional to the charge of the two objects and inversely proportional to the distance between them squared.

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What type of wave borders the violet end
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3 years ago
A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

8 0
3 years ago
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