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vazorg [7]
3 years ago
5

You want the current amplitude through a inductor with an inductance of 4.70 mH (part of the circuitry for a radio receiver) to

be 2.40 mA when a sinusoidal voltage with an amplitude of 12.0 V is applied across the inductor. What frequency is required?
Physics
1 answer:
goldenfox [79]3 years ago
3 0

Answer:

f = 1.69*10^5 Hz

Explanation:

In order to calculate the frequency of the sinusoidal voltage, you use the following formula:

V_L=\omega iL=2\pi f i L         (1)

V_L: voltage = 12.0V

i: current  = 2.40mA = 2.40*10^-3 A

L: inductance = 4.70mH = 4.70*10^-3 H

f: frequency = ?

you solve the equation (1) for f and replace the values of the other parameters:

f=\frac{V_L}{2\pi iL}=\frac{12.0V}{2\pi (2.4*10^{-3}A)(4.70*10^{-3}H)}=1.69*10^5Hz      

The frequency of the sinusoidal voltage is f

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two skateboarders of mass 50 kg and 60 kg push each other with force 70N.what is the acceleration of each skaters.step by step.
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Answer:

f=ma

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m1=50kg

m2= 60kg

a1= f/m

a1= 70/50

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6 0
3 years ago
When a driver presses the brake pedal, his car can stop with an acceleration of -5.4m/s^2. How far will the car travel while com
Dahasolnce [82]
Information that is given:
a = -5.4m/s^2
v0 = 25 m/s
---------------------
S = ?
Calculate the S(distance car traveled) with the formula for velocity of decelerated motion:
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The velocity at the end of the motion equals zero (0) because the car stops, so v=0.
0 = v0^2 - 2aS
v0^2 = 2aS
S = v0^2/2a
S = (25 m/s)^2/(2×5.4 m/s^2)
S = (25 m/s)^2/(10.8 m/s^2)
S = (625 m^2/s^2)/(10.8 m/s^2)
S = 57.87 m
3 0
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igomit [66]

Answer;

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Explanation;

"from earth to moon" implies endpoints at both locations and it is thus a line segment

A line extends forever in both directions, a line segment is just part of a line. It has two endpoints, and a ray starts at one point and continues on forever in one direction.

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Answer:

B. Solids, liquids, and gases.

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