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Ad libitum [116K]
3 years ago
6

During a golf drive, the angular velocity of the driver is 20 rad/s just before impact with the golf ball. If the distance from

the club head to the axis of rotation is 2.0 m, what is the linear velocity of the club head?
Physics
1 answer:
Julli [10]3 years ago
8 0

Answer:

Linear velocity of the club head is 40 m/s

Explanation:

To calculate the linear velocity of the club head, we just need to multiply the angular velocity of the driver (Av = 20 rad/s) by the distance from the club head to the axis of rotation (d = 2.0 m)

So, the linear velocity of the club head (Lv) is:

Lv = Av * d = 20 * 2 = 40 m/s

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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
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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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