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gladu [14]
3 years ago
8

When I move a conductor through a magnetic field, the change in magnetic field creates a net change in magnetic flux. The EMF th

at is induced by this changing flux creates tiny motional currents that are called _______. Eddy Currents Generating Currents Faraday Currents Lenz Currents Quantizing Currents
Physics
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

Eddy currents

Explanation:

When a magnetic flux linked across the coil changes, an induced emf is developed in the coil and hence the induced current is also developed in the coil.

If the induced current in the large conductor, the current is called eddy current. It produces the heating effect.

There are several applications of the eddy current.

(1) Electromagnetic brakes

(2) Induction furnace, etc.

To minimize the effect of eddy current we use choke coils, in which the core is made of laminated thin sheets.

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4 0
3 years ago
What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?
zubka84 [21]

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

6 0
3 years ago
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A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
What is 6.9 × 107 written in standard form?
n200080 [17]
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7 0
2 years ago
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A shovel is the third class lever​
Nookie1986 [14]

Answer:

Yes

Explanation:

In a third-class lever, the effort force lies between the resistance force and the fulcrum. Some kinds of garden tools are examples of third-class levers. When you use a shovel, for example, you hold one end steady to act as the fulcrum, and you use your other hand to pull up on a load of dirt.

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