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Elenna [48]
2 years ago
10

A 100-turn coil has a radius of 7.50 cm and a resistance of 50.0 W. At what rate must a perpendicular magnetic field change to p

roduce a current of 5.00 A in the coil
Physics
1 answer:
Furkat [3]2 years ago
3 0

Answer:

dB / dt = -141.47 T / s

Explanation:

For this exercise let's use Faraday's law to calculate the induced electromotive force

          E = - N dФ / dt

the magnetic flux is

         Ф = B. A

in this case the direction of the field and the normal to area are parallel so the scalar product is reduced to the algebraic product

         Ф = B A

  we substitute

        E = - N A dB / dt

the area of ​​the loop is

        A = π r²

we substitute

          E = - N π r² dB / dt

in the exercise indicate that the resistance of the coil is R = 50.0 Ω,

           E = i R

we substitute

           i R = -N π r² dB / dt

           dB / dT = - i R / N π r²

let's calculate

           dB / dt = - 5.00 50.0 / (100 π 0.075²)

           dB / dt = -141.47 T / s

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drek231 [11]

Answer:

v_f = 15 \frac{m}{s}

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum \vec{L} is

\vec{L}  = \vec{r} \times \vec{p}

where \vec{r} is the position and \vec{p} the linear momentum.

We also know that the torque is

\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )

\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p}

\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F}

but, as the linear momentum is \vec{p} = m \vec{v} this means that is parallel to the velocity, and the first term must equal zero

\vec{v} \times \vec{p}=0

so

\vec{\tau} =   \vec{r} \times \vec{F}

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

\vec{\tau}_{rod} =   0

this means, for the angular momentum measure from the rod:

\frac{d\vec{L}_{rod}}{dt} =   0

that means :

\vec{L}_{rod} = constant

So, the magnitude of initial angular momentum is :

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

and the radius is 0.250 m and the mass is 2.00 kg

| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

8 0
3 years ago
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Vlad1618 [11]

Answer:

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Explanation:

From the question,

We apply newton's second law of motion

F = m(v-u)/t.................... Equation 1

Where F = force exert by the brake, v = final speed, u = initial speed m = mass of the bicycle, t = time.

make t the subject of the equation

t = m(v-u)/F................... Equation 2

Given: m = 180 kg, u = 6.0 m/s, v = 0 m/s (comes to stop), F = -1600 N ( agianst the dirction of motion)

Substitute these value into equation 2

t = 180(0-6.0)/-1600

t = -1080/-1600

t = 0.0675 seconds.

8 0
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mote1985 [20]
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angle = 180 - 51 - 22 = 107 degrees
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c^2 = a^2 + b^2 - 2ab cos C
c^2 = 76^2 + 123^2 - 2 ( 76) ( 123) cos ( 107)
c = 162.4 mi <span>the crew fly to go directly to the field
</span>
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Andru [333]
Your answer has to be B
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3 years ago
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sineoko [7]

Answer:

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Explanation:

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