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tresset_1 [31]
3 years ago
13

A loop of wire lays flat on a horizontal surface in an area with uniform magnetic field directed vertically down. Magnetic field

suddenly grows stronger. While magnetic field is changing, the induced current in the loop as viewed from above is:
A. clockwise
B. counterclockwise
C. zero
D. There is not enough information to answer this question
Physics
1 answer:
zimovet [89]3 years ago
4 0

Answer:

B. counterclockwise

Explanation:

We shall apply Lenz's law to find direction of induced current . According to it , direction of induced current is such that the magnetic field produced  by it will try to nullify the change in  magnetic field that produces it. Increase in magnetic field in down ward direction is producing it so to nullify it , magnetic field induced will be in upward direction.  For it current induced will be anticlockwise.

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If an object starts from rest, what is its initial velocity?
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Answer:

0 m/s

Explanation:

velocity= change in displacement/ time

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0/ t

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Why doesn't the force of gravity change the speed of a bowling ball as it rolls along a bowling lane?
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Because the gravitational force, which points downward, is perfectly balanced by the normal reaction of the floor of the bowling lane, which points upward. The two forces are equal in magnitude, so the net force acting vertically on the bowling ball is zero, therefore there is no acceleration along this direction. Moreover, since the ball is moving in the horizontal direction, the gravitational force has no component along this direction, so it does not change the velocity of the ball.
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Why is a warm, tropical cumulus cloud more likely to produce precipitation than a cold, stratus cloud?
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3 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
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Assoli18 [71]
A magnet has a South Pole and a North Pole. South Pole and South Pole can't connect to her other, same as North and North. The same poles push each other away.
South Pole and North Pole connect.
4 0
3 years ago
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