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raketka [301]
3 years ago
7

In an electricity experiment, a 1.10 g plastic ball is suspended on a 56.0 cm long string and given an electric charge. A charge

d rod brought near the ball exerts a horizontal electrical force F elec on it, causing the ball to swing out to a 21.0 degree angle and remain there.
What is the tension in the string?

Physics
1 answer:
Shalnov [3]3 years ago
8 0

Answer:

Tension, T = 0.0115 N                      

Explanation:

Given that,

Mass of the plastic ball, m = 1.1 g

Length of the string, l = 56 cm

A charged rod brought near the ball exerts a horizontal electrical force F on it, causing the ball to swing out to a 21.0 degree angle and remain there. According to attached figure :

T\cos\theta=mg

T is tension in the string

T=\dfrac{mg}{\cos\theta}\\\\T=\dfrac{1.1\times 10^{-3}\times 9.8}{\cos(21)}\\\\T=0.0115\ N

So, the tension in the string is 0.0115 N.

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1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
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(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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A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid
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