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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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Answer : B.  0 m/s∧2

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From the graph we see that at the time interval 5 secs,the distance travelled is 25 m. Hence the velocity (v2) =25/5 = 5m/s.

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2 years ago
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A cannon fires a cannonball 500.0 m downrange when set at a 45.0o angle. At what velocity does the cannonball leave the cannon?
kobusy [5.1K]

Answer:

v = 70 m/s

Explanation:

Range on the cannon ball is given as

d = 500.0 m

here the angle of the projection of the ball is given as 45 degree

now we know that if the velocity of the ball is "v" then its two components will be given as

v_x = vcos45

v_y = vsin45

so here time of flight of the motion is given as

T = \frac{2v_y}{g}

T = \frac{2vsin45}{g}

also the range is given as

R = v_x T

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now plug in all data in this equation

500.0 = \frac{v^2(2sin45cos45)}{g}

v = 70 m/s

4 0
3 years ago
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Ksivusya [100]

Answer:

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5 0
2 years ago
1. A step-up transformer increases 15.7V to 110V. What is the current in the secondary as compared to the primary? Assume 100 pe
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1. I_2 = 0.14 I_1

Explanation:

We have:

V_1 = 15.7 V voltage in the primary coil

V_2 = 110 V voltage in the secondary coil

The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal

P_1 = P_2\\V_1 I_1 = V_2 I_2

where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find

\frac{I_2}{I_1}=\frac{V_1}{V_2}=\frac{15.7 V}{110 V}=0.14

which means that the current in the secondary coil is 14% of the value of the current in the primary coil.

2. 5.7 V

We can solve the problem by using the transformer equation:

\frac{N_p}{N_s}=\frac{V_p}{V_s}

where:

Np = 400 is the number of turns in the primary coil

Ns = 19 is the number of turns in the secondary coil

Vp = 120 V is the voltage in the primary coil

Vs = ? is the voltage in the secondary coil

Re-arranging the formula and substituting the numbers, we find:

V_s = V_p \frac{N_s}{N_p}=(120 V)\frac{19}{400}=5.7 V

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