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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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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
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Answer:

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(b)  Electric field at 2.90 m is zero.

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Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

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