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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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Answer:

230.26 N

Explanation:

Since the speed is constant, acceleration is zero hence the net force will be given by the product of mass, coefficient of friction and acceleration due to gravity

F=0.72*32.6*9.81=230.26 N

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3 years ago
Which of the following is required for work to be done on an object?
ad-work [718]
I think that in order for work to be done, the object must move in the direction of the force and move over a distance.
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2 years ago
Which electromagnet will produce the strongest magnetic force? (1 point)
Contact [7]

Answer:

Y

Explanation:

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Have a great day :)

7 0
3 years ago
Read 2 more answers
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
2 years ago
Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s,
SSSSS [86.1K]

Answer:

Time period between the successive beats will be 0.1703 sec

Explanation:

We have given speed of the sound v = 349 m/sec

Wavelength of piano A\lambda _A=0.766m

Wavelength of piano  B\lambda _B=0.776m

So frequency of piano A f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz

Frequency of piano B f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec

So time period between the successive beats will be 0.1703 sec

4 0
3 years ago
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