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AleksAgata [21]
3 years ago
5

Suppose that the emf from a rod moving in a magnetic field was used to supply the current to illuminate a light bulb in a circui

t, and the force needed to keep the rod moving is . What happens to the force needed drive the motion of the rod if the light bulb is removed
Physics
1 answer:
faltersainse [42]3 years ago
8 0

Explanation:

The rod moving in a magnetic field and induces an emf which is used to illuminate the bulb. But if the bulb is removed form the circuit, the circuit is opened.

For an open circuit, no current is passes through the moving rod. If there is no current along the rod, then no magnetic field developed around the rod. Because moving charges nothing but current produces the magnetic field around the rod.

The formula for the magnetic force on the rod is,

F_{\mathrm{B}} &=I(l \times B)

=I l B \sin \theta

The current along the rod is zero because the bulb is removed and the magnetic field around the rod is zero because no current is passes through the rod.

Then calculate the magnetic force on the rod as follows:

\F_{\mathrm{B}} &=I l B \sin \theta

=(0)(l)(0) \sin \theta =0 \mathrm{N}

Thus, no force is needed because there is no longer magnetic field developed around the rod.

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7 0
3 years ago
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A circuit contains an EMF source, a resistor R, a capacitor C, and an open switch in series. The capacitor initially carries zer
Flura [38]

Answer:

t = 1.098*RC

Explanation:

In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:

Q=Q_{max}[1-e^{-\frac{t}{RC}}]         (1)

Qmax: maximum charge capacity of the capacitor

t: time

R: resistor of the circuit

C: capacitance of the circuit

When the capacitor has 2/3 of its maximum charge, you have that

Q=(2/3)Qmax    

You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:

Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC

The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC

6 0
4 years ago
Topic: Chapter 19: Some wiggle room
Nana76 [90]

Answer:

4.86 m

Explanation:

Given that,

The frequency produced by a humming bird, f = 70 Hz

The speed of sound, v = 340 m/s

We need to find how far does the sound  travel between wing flaps. Let the distance is equal to its wavelength. So,

v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m

So, the sound travel 4.86 m between wings flaps.

3 0
3 years ago
A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va
ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

U=\dfrac{1}{2}CV^2

C=\dfrac{\varepsilon _oA}{d}

U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2

If energy become double U' = 2 U then d'

U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

7 0
3 years ago
At what speed must a 950-kg subcompact car be moving to have the same kinetic energy as a 3.2×104−kg truck going 20 km/h?
Ostrovityanka [42]

<u>Answer:</u>

 950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

<u>Explanation:</u>

  Kinetic energy of body with mass m and moving at velocity v is given by

                  KE=\frac{1}{2} mv^2

  KE of truck  = \frac{1}{2}*3.2*10^4*20^2

  KE of subcompact car = \frac{1}{2}*950*v^2

  Equating

         \frac{1}{2}*3.2*10^4*20^2= \frac{1}{2}*950*v^2\\ \\ v=116.08 km/hr

 So 950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

6 0
4 years ago
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