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kakasveta [241]
3 years ago
12

A voltmeter with resistance RV is connected across the terminals of a battery of emf and internal resistance r. Find the potenti

al difference Vmeter measured by the voltmeter.

Physics
1 answer:
makvit [3.9K]3 years ago
4 0

Answer:

voltage measured by the voltmeter = (E × RV)/(r + RV)

Explanation:

The circuit diagram for this description is presented in the attached image.

The internal resistance of an emf source is modelled to be in series with the source.

Therefore, the end product is a circuit with the battery in series connection with the internal resistance and resistance of the volunteer. The voltage picked up by the volunteer is the voltage across resistor RV.

Total resistance in the circuit = (r + RV) ohms (since both resistors are in parallel)

Current produced by the emf source = E/(Total resistance) = E/(r + RV)

The voltage across resistor RV = current flowing through this resistor × its resistance.

Since all the circuit elements are in series with each other, same current, E/(r + RV) flows through them all

Voltage across RV = voltage measured by the voltmeter = [E/(r + RV)] × RV = (E × RV)/(r + RV) = (E.RV)/(r + RV)

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A well-trained athlete can run 400m in 47s, what is the athlete’s velocity?
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8.51 m/s

Explanation:

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Velocity = 400 m ÷ 47 s

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You have a string with a mass of 0.0133 kg. You stretch the string with a force of 8.89 N, giving it a length of 1.97 m. Then, y
Kazeer [188]

Answer:

(i) The wavelength is 0.985 m

(ii) The frequency of the wave is 36.84 Hz

Explanation:

Given;

mass of the string, m = 0.0133 kg

tensional force on the string, T = 8.89 N

length of the string, L = 1.97 m

Velocity of the wave is:

V = \sqrt{\frac{F_T}{M/L} } \\\\V = \sqrt{\frac{8.89}{0.0133/1.97} } \ = 36.29 \ m/s

(i) The wavelength:

Fourth harmonic of a string with two nodes, the wavelength is given as,

L = 2λ

λ = L/2

λ = 1.97 / 2

λ = 0.985 m

(ii) Frequency of the wave is:

v = fλ

f = v / λ

f = 36.29 / 0.985

f = 36.84 Hz

3 0
3 years ago
2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across
gulaghasi [49]

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

hence, force between 3 kg and 5 kg block = (16-6) = 10N .

answers:-

(a) 20 N

(b) 16N

(c) 10 N

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1 year ago
Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with e
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To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.

However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that

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k = Boltzmann constant

V = Volume

T = Temperature

P = Pressure

Our values are given as,

N = 1

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Therefore the Length would be given as,

L = V^{1/3}

L = (4.0858*10^{-26})^{1/3}

L = 3.445*10^{-9}m

Therefore each length of the cube is 3.44nm

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3 years ago
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