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PolarNik [594]
3 years ago
15

The acceleration due to gravity on the Moon is gM. Suppose an astronaut on the Moon drops an object from a height of H. The time

it would take the object to reach the Moon’s surface would be TM. The same object is dropped from the same height on Earth, where the acceleration due to gravity is gE. The time it takes the object to reach the Earth’s surface is TE. Which of the following is a correct mathematical relationship for the two times?
TE=gEgM−−−√TM

TE=gEgMTM
A

TE=gMgE−−−√TM

TE=gMgETM
B

TE=gEgMTM

TE=gEgMTM
C

TE=gMgETM

TE=gMgETM
D

TE=TM
Physics
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

TE = sqrt(GM/GE)TM

Explanation:

To solve for this problem, you have to use the second kinematic equation and set the height equal to each other.  Because the heights are equal, 1/2GETE^2 = 1/2GMTM^2.  Rearrange the equation and you'll get the answer

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Answer:

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Explanation:

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weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

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Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

Length of the inclined edge = 16 m

v^{2}=u^{2}+2as

v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

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4 years ago
What is the value of acceleration due to gravity at heght h=4re​
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Explanation:

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valentina_108 [34]

Answer: Diagram B

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Find the voltage across the 15 Q resistor.<br> [?] V<br><br> No links please
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Answer:

Explanation:

same idea as before Liam,  first, find the parallel resistance in 35 || 20

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V = IR

10 = I *    27.727272

10 / 27.727272 = I

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