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lara [203]
3 years ago
11

A magnetic field is passing through a loop of wire whose area is 0.020 m2. The direction of the magnetic field is parallel to th

e normal to the loop, and the magnitude of the field is increasing at the rate of 0.24 T/s. (a) Determine the magnitude of the emf induced in the loop. (b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 2.1 T if the induced emf is to be zero? Use a minus sign if the area is to be shrunk
Physics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

a) E = 0.0048 Volts

b)  dA/dt = - 0.002285 m²/s

Explanation:

Given:

Area, A = 0.020 m²

Rate of change of magnetic field, dB/dt = 0.24 T/s

a) The magnitude of the emf induced (E) is given as:

  E= A × (dB/dt)

  on substituting the values in the above equation, we get

E = (0.020 m²) × (0.24 T/s)

or

E = 0.0048 Volts

b) Now, The induced emf when both the area and the magnetic field is varying

we have

E = B(dA/dt) + A(dB/dt)

Now, for the given case induced emf is zero i.e E = 0 and magnetic field B = 2.1 T

thus,

0 = (2.1 T)(dA/dt) + (0.020 m2)(0.24 T/s)

dA/dt = - 0.002285 m²/s

Hence, the area should be decreased at the rate of 0.002285 m²/s

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Mass of the object is given as

m = 1417 g = 1.417 kg

now the speed of object is given as

v = \frac{d}{t}

here we know that

d = 47 m

t = 90 s

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KE = \frac{1}{2}(1.417)(0.52)^2

KE = 0.19 J

now the power is defined as rate of energy

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P = \frac{KE}{t}

P = \frac{0.19}{90} = 2.14\times 10^{-3} W

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3 years ago
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4 0
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IRISSAK [1]

It's just asking you to sit down and COUNT the little squares in each sector.

It'll help you keep everything straight if you take a very sharp pencil and make a tiny dot in each square as you count it.  That way, you'll be able to see which ones you haven't counted yet, and also you won't count a square twice when you see that it already has a dot in it.

(If, by some chance, this is a picture of the orbit of a planet revolving around the sun ... as I think it might be ... then you should find that both sectors jhave the same number of squares.)  

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The atmosphere of Neptune and Uranus have a blue color because of which gas?
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oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
puteri [66]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

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3 years ago
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