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mart [117]
4 years ago
15

If the rate of change of the magnetic field applied to a loop of wire is doubled, what happens to the induced emf in that loop a

ssuming that all other parameters remain unchanged?
Physics
1 answer:
slava [35]4 years ago
3 0

Answer:

It is doubled.

Explanation:

The EMF( Electromotive force) is usually gotten from an energy source. In this case it is from the magnetic field. However when the other parameters are constant then the main focus is between the EMF and the magnetic field. They have a direct proportion relationship which is why when the magnetic field is doubled the EMF is doubled too.

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In which regions can the gravitational field strength due to the two planets be zero? Explain.
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Answer:

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Explanation:

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3 years ago
You heat a mug of water to make hot chocolate. Which statement best describes the changes in the water?
PIT_PIT [208]

It is A because the hot chocolate is well, ya know, hot..

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This is a change in an object's position. <br> Net Force<br> Gravitational Pull<br> Motion<br> Force
alexira [117]

Answer:

Motion

Explanation:

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3 years ago
An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers
ohaa [14]

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E     where E is the value of the field and N e the charge Q

M g = N e E      and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check:     N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13   electrons

7 0
3 years ago
A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position
Illusion [34]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Place of particles

X(t)=\int V_{x}(t)dt

       =\int 2t^{2}dt\\\\=\frac{2}{3}t^{3}+C

\to t=0\\\\ \to X(0)=2.3 \ m

\to X(0)=0+C\\\\ \to C=2.3\  m

\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3 \\\\

        = \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\  = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m

In point b:

when t=2.2 \ s

the Particle velocity  (V)=2 \times 2.22 =9.68\  \frac{m}{s}

In point c:

Calculating the Particle acceleration:

\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2  =8.8 \ \frac{m}{s^2}

8 0
3 years ago
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