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Elza [17]
3 years ago
8

HELP PLS MARKING BRANLIST 100 Pts TAKING TEST RN

Physics
2 answers:
FromTheMoon [43]3 years ago
6 0

Answer:

A. 1.90

I divided 63 by 60 and got 1.05

I multiplied 1.05 by 2 (120 seconds = 2 minutes) and I got 2.1

1.90 is the closest answer.

Bumek [7]3 years ago
5 0

Answer:

It will travel 7,560 m

Explanation:

63 x 120 = 7,560 m

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Four satellites are in orbit around the Earth. The heights and the masses of
Kaylis [27]

Answer:

satellite B

Explanation:

A .F= G (mM)/r²

B .F= G (2mM)/r² = 2G (Mm)r²

C .F= G (3mM)/(2r)² = ¾G (mM)/r²

D .F= G (4mM)/(2r)² = G (mM)/r²

4 0
3 years ago
Which of the following describes sound waves?
lianna [129]
The answer is c) electromagnetic sawed in which the vibrations are perpendicular to the motion of the sound
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Which of these statements is correct about Newton's first law of motion?
nexus9112 [7]

Answer:

Answer D: It describes the relationship between motion and force.

Explanation:

The answer is D because a law is something that describes something in nature, but does not try to explain how or why is occurs (that is a theory). Options B and C sound more like theories, while option A sounds like a definition. Option D is correct because a law describes without explaining.

4 0
3 years ago
(b) If you decrease the length of the pendulum by 25%, how does the new period TN compare to the old period T?
malfutka [58]

Answer:

The new period will be reduced by 50%

Explanation:

The period of pendulum is given by;

T= 2\pi\sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g} }\\\\(\frac{T}{2\pi} )^2 = {\frac{L}{g}}\\\\\frac{T^2}{4\pi ^2} = {\frac{L}{g}}\\\\T^2(\frac{g}{4\pi ^2}) = L\\\\ \frac{g}{4\pi ^2}= \frac{L}{T^2}\\\\\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}

When the length is decreased by 25%, the new length L₂ is given by;

L₂ = 25/100(L₁)

L₂ = 0.25L₁

\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{T_1^2L_2}{L_1} \\\\T_N^2 = \frac{T^2(0.25L_1)}{L_1}\\\\ T_N^2 =0.25T^2\\\\T_N = \sqrt{0.25T^2}}\\\\T_N = 0.5 T

Thus, the new period will be reduced by 50%

8 0
3 years ago
A 2.10 X 10^3 kg car starts from rest at the top of a 5.0
xeze [42]

Answer:

3.80 m/s

Explanation:

Solution: From the question we have:

A 2.10 x 10^3 kg car starts from rest at the top of a 5.0 m long driveway that is sloped at 20 degrees with the horizontal.

average friction force of 4.0 x 10^3 N impedes the motion

To find out:  find the speed of the car at the bottom of the driveway.

=>h = 5×sin20 = 1.71 m.

PE = mgh = 2100×9.8×1.71 = 35,192 J.

KE = PE-work done by frictional force

KE=PE-4000×5

0.5mV^2 = 35192 - 4000×5

1050V^2 = 15,192

V^2 = 14.47

V = 3.80 m/s.

4 0
3 years ago
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