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Rufina [12.5K]
4 years ago
10

A loop of wire with cross-sectional area 1 m2 is inserted into a uniform magnetic field with initial strength 1 T. The field is

parallel to the axis of the loop. The field begins to grow with time at a rate of 2 Teslas per hour. What is the magnitude of the induced EMF in the loop of wire
Physics
1 answer:
iragen [17]4 years ago
8 0

Answer:

The magnitude of the EMF is 0.00055  volts

Explanation:

The induced EMF is proportional to the change in magnetic flux based on Faraday's law:

emf\,=-\,N\, \frac{d\Phi}{dt}

Since in our case there is only one loop of wire, then N=1 and we get:

emf\,=-\,N\, \frac{d\Phi}{dt}

We need to express the magnetic flux given the geometry of the problem;

\Phi=B\,\,Awhere A is the area of the coil that remains unchanged with time, and B is the magnetic field that does change with time. Therefore the equation for the EMF becomes:

emf\,=-\,N\, \frac{d\Phi}{dt} = \frac{d\Phi}{dt} =-\frac{d\,(B\,A)}{dt} =-\,A\,\frac{d\,(B)}{dt}=- 1\,m^2(2\,\,T/h})= -2\,\,m^2\,T/(3600\,\,s)= -0.00055\,Volts

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A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of 25 m along t
Rus_ich [418]

Answer:

The centripetal acceleration of the girl is 2.468 m/s²

Explanation:

Given;

number of turns, = ¹/₄ Revolution

distance traveled by the girl, d = 25 m

time of motion, t = 5.0 s

The linear speed of the of the girl is calculated as;

v = \omega \ r\\\\v =(\frac{1}{4}rev \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1}{5 \ s} )  (25 \ m)\\\\v =  (0.3142 \ \frac{rad}{s} )(25 \ m)\\\\v = 7.855 \ m/s

The centripetal acceleration of the girl is calculated as;

a_c = \frac{v^2}{r} \\\\a_c = \frac{(7.855)^2}{25} \\\\a_c = 2.468 \ m/s^2

Therefore, the centripetal acceleration of the girl is 2.468 m/s²

8 0
3 years ago
13. An object with a mass of 2.0 kg has a force of 4.0 newtons applied to it.
julsineya [31]

Answer:2m/s²

Explanation: Well F=MA so sice F=4N and M=2kg let's plug in the values

4N=2KG*A

A=4N/2KG

A=2m/s²

7 0
4 years ago
In a double slit experiment, the intensity of light at the center of the central bright fringe is measured to be 6.2 µW/m2. What
-BARSIC- [3]

Answer:

   I_FWHW = 3.2 μW / m²

Explanation:

In the analysis of optics and electricity a very useful magnitude is the width at half height (FWHW) and the intensity at this height, which is given by

               I_FWHW = I₀ / 2

corresponds to the width of the line for this intensity.

In this case they give the maximum intensity for which

               I_FWHW = 6.2 / 2

               I_FWHW = 3.2 μW / m²

You do not give more data in your exercise, but the most interesting calculation is to find the angle values ​​for which you have this intensity since it is this range is 50% of the energy of the system, have I write the equation for this calculation

             I = Io cos² x₁   (sin x / x)²

             x₁ = π d sin θ /λ

             x = π b sin θ /λ

where d is the separation of the slits and b the width of each slit

5 0
4 years ago
How does heat affect the thermal energy of an object that is colder than the qir
podryga [215]

Answer:

the correct answer is B

Explanation:

thermal energy makes heat increase.

4 0
3 years ago
Three identical balls are thrown from the same height off a tall building. They are all thrown with the same speed, but in diffe
koban [17]

Answer:

three balls have the same speed

Explanation:

In a parabolic motion we have that

v_{x}=v_{0}cos\alpha\\v_{y}=-gt+v_{0}sin\alpha\\v=\sqrt{v_{x}^{2}+v_{y}^{2}}\\

but the time just before the balls hit the ground is

t=\frac{2v_{0}sin\alpha}{g}

Hence we have

ball A

v=\sqrt{v_{0}^{2}cos^{2}(0)+(-g\frac{2v_{0}sin(0)}{g}+v_{0}sin(0))^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}

ball B

v=\sqrt{v_{0}^{2}cos^{2}(45)+(-g\frac{2v_{0}sin(45)}{g}+v_{0}sin(45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}

ball C

v=\sqrt{v_{0}^{2}cos^{2}(-45)+(-g\frac{2v_{0}sin(-45)}{g}+v_{0}sin(-45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}

Hence, all three balls have the same speed just before hit the groug

vA=vB=vC

I hope this is useful for you

regards

5 0
3 years ago
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