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Ivan
2 years ago
8

Three joules of work is needed to shift 10 C of charge from one place to another. The potential difference between the places is

______. *
0.3 V
0.5 V
5 V
3 V​
Physics
1 answer:
dimaraw [331]2 years ago
4 0

Answer:

The potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

Explanation:

Given

Work done W = 3J

Amount of Charge q = 10C

To determine

We need to determine the potential difference V between the places.

The potential difference between the two points can be determined using the formula

Potential Difference (V) = Work Done (W) / Amount of Charge (q)

or

\:V\:=\:\frac{W}{q}

substituting W = 3 and q = 10 in the formula

V=\frac{3}{10}

V=0.3 V

Therefore, the potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

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The first person = A

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7 0
3 years ago
A sound from a source has an intensity of 270 dB when it is 1 m from the source.
Rufina [12.5K]
Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: \frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2
where
I_{2} is the intensity at distance 2
I_{1} is the intensity at distance 1
d_{2} is distance 2
d_{1} is distance 1

We can infer for our problem that I_{1}=270, d_{1}=1, and d_{2}=3. Lets replace those values in our formula to find I_{2}:
\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2
\frac{I_{2} }{270} =( \frac{1}{3} )^2
\frac{I_{2} }{270} = \frac{1}{9}
I_{2}= \frac{270}{9}
I_{2}=30 dB

We can conclude that the intensity of the sound when is <span>3 m from the source is 30 dB.</span>
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3 years ago
What unit is used to measure loudness and at what level can hearing loss happen?
valkas [14]
Decibel<span> (dB) and, hearing loss levels can be anywhere like </span><span><span>-56 to -70 dB for Moderate/severe hearing loss, -</span><span>71 to -90 dB for Severe loss and</span><span> -91 dB for <span>Profound loss.</span></span></span>
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3 years ago
How do you calculate speed
Kryger [21]

Answer:

distance(d)=.....

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8 0
3 years ago
Read 2 more answers
A motor has an armature resistance of 3.75 Ω . Part A If it draws 9.10 A when running at full speed and connected to a 120-V lin
bulgar [2K]

Back emf is 85.9 V.

<u>Explanation:</u>

Given-

Resistance, R = 3.75Ω

Current, I = 9.1 A

Supply Voltage, V = 120 V

Back emf = ?

Assumption - There is no effects of inductance.

A motor will have a back emf that opposes the supply voltage, as the motor speeds up the back emf increases and has the effect that the difference between the supply voltage and the back emf is what causes the current to flow through the armature resistance.

So if 9.1 A flows through the resistance of 3.75Ω then by Ohms law,

The voltage across the resistance would be

v = I x R

  = 9.1 x 3.75

  = 34.125 volts

We know,

supply voltage = back emf + voltage across the resistance

By plugging in the values,

120 V = back emf + 34.125 V

Back emf = 120 - 34.125

                = 85.9 Volts

Therefore, back emf is 85.9 V.

4 0
3 years ago
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