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2 years ago

Which law says that the induced voltage is directly proportional to the magnetic flux?

Physics

2 answers:

Vsevolod [243]2 years ago

7 0

law of electromagnetic induction hope this helps

LUCKY_DIMON [66]2 years ago

7 0

It is Faraday's law, so if your doing this from plato it should be answer B, hope this helps.

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When the pendulum bob reaches the mean position, the net force acting on it is zero. Why then does it swing past the mean positi

ryzh [129]

Answer:

<u>The pendulum bob swing past the mean position because:</u>

When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to the restoring toque the bob starts to accelerates towards its mean postion. The maximum acceleration of the pendulum's bob is $-w^{2} A$ and the the acceleration decreases as $-w^{2} x$ towards the mean position.

The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.

6 0

2 years ago

What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal

galben [10]

Hello

1) First of all, since we know the radius of the wire ( $r=1.2~mm=0.0012~m$ ), we can calculate its cross-sectional area
$A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2$

2) Then, we can calculate the current density J inside the wire. Since we know the current, $I=3~A$ , and the area calculated at the previous step, we have
$J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2$

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity $\sigma=3.6\cdot10^7~ \frac{A}{Vm}$ of the aluminium, the electric field is given by
$E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m$

4 0

3 years ago

A student carries a very heavy backpack. To lift the backpack off the ground, the student must apply 80 N of force to do so. The

Xelga [282]

The work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.

<h3>How to calculate work done?</h3>

Work done is a measure of energy expended in moving an object; most commonly, force times distance.

It is said that no work is done if the object does not move, hence, the work done on an object can be calculated as follows:

Work done = Force × Distance

According to this question, a student carries a very heavy backpack and to lift the backpack off the ground, the student must apply 80 N of force to lift the backpack 1.5 m.

Work done = 80N × 1.5m

Work done = 120J

Therefore, the work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.

Learn more about work done at: brainly.com/question/28172139

#SPJ1

8 0

1 year ago

What does a line segment with a zero slope from a position-time graph indicate about the motion of the object in that interval o

Blizzard [7]

the object is not moving

3 0

2 years ago

Charge Q = +4.00 mC is distributed uniformly over the volume of an insulating sphere that has radius R = 5.00 cm. What is the po

sattari [20]

Answer:

3.6 x 10^8 V

Explanation:

Q = 4 m C = 4 x 10^-3 C

r = 5 cm = 0.05 m

The formula for the potential at the surface is

Vs = K Q / r = (9 x 10^9 x 4 x 10^-3) / 0.05 = 7.2 x 10^8 V

The formula for the potential at the centre is

Vc = 3/2 Vs

Vc = 1.5 x 7.2 x 10^8 V = 10.8 x 10^8 V

The difference in potential is

V = Vc - Vs = 10.8 x 10^8 - 7.2 x 10^8 = 3.6 x 10^8 V

8 0

2 years ago