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Mila [183]
3 years ago
10

A car accelerates uniformly in a straight line

Physics
1 answer:
julia-pushkina [17]3 years ago
8 0

The car travels a distance <em>d</em> from rest with acceleration <em>a</em> after time <em>t</em> of

<em>d</em> = 1/2 <em>a</em> <em>t</em>²

It covers 69 m with 2.8 m/s² acceleration, so that

69 m = 1/2 (2.8 m/s²) <em>t</em>²

<em>t</em>² = 2 (69 m) / (2.8 m/s²)

<em>t</em> ≈ 7.02 s

where we take the positive square root because we're talking about time *after* the car begins accelerating.

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A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long an
LUCKY_DIMON [66]

Answer:

a) 1.34*10^-8 W

b) 1.18*10^-5 H

c) 20mV

Explanation:

a) To find the average magnetic flux trough the inner solenoid you the following formula:

\Phi_B=BA=\mu_oNIA

mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

N: turns of the solenoid = 340

I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m

You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

M=\mu_o N_1 N_2 \frac{A_2}{l}

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H

the mutual inductance is 1.18*10^{-5}H

c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:

\epsilon_1=M\frac{dI_2}{dt}

by replacing you obtain:

\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV

the emf is 20mV

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