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elena-s [515]
3 years ago
8

The current in a coil with a self inductance of 1.5 mH increases from 0 to 1.0 A in a tenth of a second. What is the induced emf

in the coil
Physics
1 answer:
krok68 [10]3 years ago
3 0

Answer:

the induced emf in the coil is 15 mV.

Explanation:

Given;

self inductance, L = 1.5 mH = 1.5 x 10⁻³ H

change in current = dI = 1.0 A

change in time, dt = 1/10 = 0.1

The induced emf is calculated as;

Induced \ EMF = L (\frac{dI}{dt} )\\\\Induced \ EMF = 1.5  \times  10^{-3}\ (\frac{1}{0.1} )\\\\Induced \ EMF = 1.5  \times  10^{-3} \ \times \ 10\\\\Induced \ EMF = 15  \times  10^{-3} \ V\\\\Induced \ EMF = 15 \ mV

Therefore, the induced emf in the coil is 15 mV.

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A 24-cm-diameter vertical cylinder is sealed at the top by a frictionless 15 kg piston. The piston is 90 cm above the bottom whe
abruzzese [7]

Answer:

(A) P_i=3249.41\ Pa

(B) V_f=0.3358\ m

Explanation:

Given:

  • diameter of the cylinder, d= 0.24\ m
  • mass of piston sealing on the top, m_p=15\ kg
  • initial temperature of the piston, T_i=315+273= 588\ K
  • initial height of piston, h_i=0.9\ m
  • atmospheric pressure on the piston, p_a=1 atm=101325\ Pa

(A)

<u>Initial pressure of gas is the pressure balanced by the weight of piston:</u>

P_i=\frac{m_p.g}{\pi.d^2\div 4}

P_i=\frac{15\times 9.8}{\pi\times (0.24^2\div 4)}

P_i=3249.41\ Pa

<em>Which is gauge pressure because it is measured with respect to the atmospheric pressure.</em>

(B)

Given:

  • Final temperature, T_f=18+273=291\ K

<u>Now, volume of air initially in the cylinder:</u>

V_i=\pi.d.h_i

V_i=\pi\times 0.24\times 0.9

V_i=0.6786\ m^3

Using gas law:

\frac{P_i V_i}{T_i}= \frac{P_f V_f}{T_f} ........................................(1)

<em>∵In every condition of equilibrium the gas pressure will be balanced by the weight of the piston so it is an </em><em>isobaric transition</em><em>.</em>

∴P_i=P_f

<u>Hence eq. (1) is reduced to:</u>

\frac{V_i}{T_i}= \frac{V_f}{T_f}

putting respective values:

\frac{0.6786}{588}= \frac{V_f}{291}

V_f=0.3358\ m

6 0
4 years ago
How much power does it take to lift 30.0 N 10.0 m high in 10.00 s?
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The power required is 30 Watt.

Let us recall that power is defined as the rate of doing work. Hence, we can write as follows;

Power = Work done/ time taken

Now;

work done =  Force × distance

Force = 30.0 N

Distance = 10.0 m

work done = 30.0 N × 10.0 m = 300 J

The power expended = 300 J/10.00 s = 30 Watt

Learn more: brainly.com/question/64224

8 0
3 years ago
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If the treadmill’s initial speed is 1.7m/s, what will its final speed be?
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For how long? Or is it just speed ??
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What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight ou
Lubov Fominskaja [6]

Answer:

The torque about his shoulder is 34.3Nm.

The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.

Explanation:

The solution to the problem can be found in the attachment below.

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A 50 kg car is pushed so that it speeds up from 20 m/s to 50 m/s in 3 seconds. What is the force acting on the car?
egoroff_w [7]

Answer:

A. 500 N

Explanation:

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a = Δv / Δt

a = (50 m/s − 20 m/s) / 3 s

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Force is mass times acceleration.

F = ma

F = (50 kg) (10 m/s²)

F = 500 N

3 0
3 years ago
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