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Harman [31]
2 years ago
9

An object of mass M is dropped near the surface of Earth such that the gravitational field provides a constant downward force on

the object. Which of the following describes what happens to the center of mass of the object-Earth system as the object falls downward toward Earth? a. It moves toward the center of Earth. b. It moves toward the object.c. It does not move. d. The answer cannot be determined without knowing the mass of Earth and the distance between the object and Earth’s center.
Physics
1 answer:
marysya [2.9K]2 years ago
6 0

Answer:

The answer is: c. It does not move

Explanation:

Because the gravitational force is characterized by being an internal force within the Earth-particle system, in this case, the object of mass M. And since in this system there is no external force in the system, it can be concluded that the center of mass of the system will not move.

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Answer:

Explanation:

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v² = u² + 2gh₁

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v = 6.64 m / s

It becomes initial velocity during impact .

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This becomes final velocity after impact

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m ( final velocity - initial velocity )

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= .7056 N.s.

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Which optical device can focus light to a point through reflection?
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Identify whether the features and functions belong to the Hubble Space Telescope or the ALMA radio telescope. 1.installed on lan
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The average kinetic energy of the molecules of an ideal gas at 10∘C has the value K10. At what temperature T1 (in degrees Celsiu
Westkost [7]

Answer:

A) T1 = 566 k = 293°C

B) T2 = 1132 k = 859°C

Explanation:

A)

The average kinetic energy of the molecules of an ideal gas is givwn by the formula:

K.E = (3/2)KT

where,

K.E = Average Kinetic Energy

K = Boltzman Constant

T = Absolute Temperature

At 10°C:

K.E = K10

T = 10°C + 273 = 283 K

Therefore,

K10 = (3/2)(K)(283)

FOR TWICE VALUE OF K10:

T = T1

Therefore,

2 K10 = (3/2)(K)(T1)

using the value of K10:

2(3/2)(K)(283) = (3/2)(K)(T1)

<u>T1 = 566 k = 293°C</u>

<u></u>

B)

The average kinetic energy of the molecules of an ideal gas is given by the formula:

K.E = (3/2)KT

but K.E is also given by:

K.E = (1/2)(m)(vrms)²

Therefore,

(3/2)KT = (1/2)(m)(vrms)²

vrms = √(3KT/m)

where,

vrms = Root Mean Square Velocity of Molecule

K = Boltzman Constant

T = Absolute Temperature

m = mass

At

T = 10°C + 273 = 283 K

vrms = √[3K(283)/m]

FOR TWICE VALUE OF vrms:

T = T2

Therefore,

2 vrms = √(3KT2/m)

using the value of vrms:

2√[3K(283)/m] = √(3KT2/m)

2√283 = √T2

Squaring on both sides:

(4)(283) = T2

<u>T2 = 1132 k = 859°C</u>

8 0
2 years ago
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