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Gelneren [198K]
3 years ago
5

Why are the action and reaction forces described by Newton’s third law of motion excluded in the free-body diagram of an object?

Physics
2 answers:
Lorico [155]3 years ago
6 0

The correct answer is:

B. They act on different objects, so they would not appear together.

Let's take for example a box sitting on a table. If we draw the free-body diagram of the box, we would draw two forces:

- the weight of the box (downward)

- the normal reaction of the table on the box (upward)

However, this is not a pair of action-reaction forces as described in Newton's third law. In fact, the correct action-reaction pair would be:

- weight of the box (=gravitational force exerted by the Earth on the box)

- gravitational force exerted by the box on the Earth

We see that these two forces act on two different objects (the first one on the box, the second one on the Earth), so they do not appear in the same free-body diagram.

umka2103 [35]3 years ago
4 0
B. <span>They act on different objects, so they would not appear together.</span>
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A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×10^{4} 4 m/s^{2} 2 , and 1.85 ms (1
Paha777 [63]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 =  u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

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6 0
3 years ago
1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB

<h3>Speed of sound at the given temperature</h3>

v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )

where;

  • -v₀ is velocity of the observer moving away from the source
  • -vs is the velocity of the source moving towards the observer
  • fs is the source frequency
  • fo is the observed frequency
  • v is speed of sound

f_0 = f_s(\frac{v-v_0}{v- v_s} )

f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz

Learn more about intensity of sound here: brainly.com/question/17062836

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2 years ago
Ella makes this table to organize her notes on whether atoms gain or lose energy during the changes of state.
solong [7]

Answer:

The answer is D.

7 0
3 years ago
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Temperature which is solid becomes a liquid
Brums [2.3K]

Answer:

The answer would be melting point.

Explanation:

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5 0
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An object with a mass of 9 kg weighs 14.4 N on the Moon. What is the
erma4kov [3.2K]

Answer:

1.6 m/s²

Explanation:

Weight equals mass times acceleration due to gravity.

F = mg

14.4 N = (9 kg) g

g = 1.6 m/s²

7 0
3 years ago
Read 2 more answers
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