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PilotLPTM [1.2K]
3 years ago
10

If I wanted to generate a maximum emf of 20 V, what angular velocity (radians/sec, aka Hz) would be required given a circular co

il of wire with diameter 40 cm and 500 coils? The coil rotates through a magnet field of magnitude 9e-3 T which is directed such that the angle between the area vector and the magnet field vector varies from 0 to 2 π radians.
Physics
1 answer:
sergij07 [2.7K]3 years ago
6 0

Answer:

Angular velocity, \omega=35.36\ rad/s

Explanation:

It is given that,

Maximum emf generated in the coil, \epsilon=20\ V

Diameter of the coil, d = 40 cm

Radius of the coil, r = 20 cm = 0.2 m

Number of turns in the coil, N = 500

Magnetic field in the coil, B=9\times 10^{-3}\ T

The angle between the area vector and the magnet field vector varies from 0 to 2 π radians. The formula for the maximum emf generated in the coil is given by :

\epsilon=NBA\omega

\omega=\dfrac{\epsilon}{NBA}

\omega=\dfrac{20\ V}{500\times 9\times 10^{-3}\ T\times \pi (0.2\ m)^2}

\omega=35.36\ rad/s

So, the angular velocity of the circular coil is 35.36 rad/s. Hence, this is the required solution.

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1. A brick and a peanut are dropped of the roof of a house at the same time, which
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Answer:

It would be the brick since the mass and the weight is greater than the peanut I might be wrong on this but thats the best answer i can give

Explanation:

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3 years ago
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that
pav-90 [236]

Answer:

The permittivity of rubber is  \epsilon  = 8.703 *10^{-11}

Explanation:

From the question we are told that

     The  magnitude of the point charge is  q_1 =  70 \ nC  =  70 *10^{-9} \  C

      The diameter of the rubber shell is  d = 32 \ cm  =  0.32 \ m

       The Electric field inside the rubber shell is  E =  2500 \ N/ C

The radius of the rubber is  mathematically evaluated as

              r =  \frac{d}{2} =  \frac{0.32}{2}  =  0.16 \ m

Generally the electric field for a point  is in an insulator(rubber) is mathematically represented as

         E =  \frac{Q}{ \epsilon }  *  \frac{1}{4 *  \pi r^2}

Where \epsilon is the permittivity of rubber

    =>     E  *  \epsilon  *  4 * \pi *  r^2 =  Q

   =>      \epsilon  =  \frac{Q}{E *  4 *  \pi *  r^2}

substituting values

            \epsilon  =  \frac{70 *10^{-9}}{2500 *  4 *  3.142 *  (0.16)^2}

            \epsilon  = 8.703 *10^{-11}

7 0
3 years ago
A stone that is thrown vertically upwards was having a velocity of 15m/s after reaches 2/3 of its maximum height. What is the ma
attashe74 [19]

<em>The correct answer is option</em><em> B.</em> The maximum height that can be reached by the stone is determined as 11.5 m.

<h3>Maximum height attained by the stone </h3>

The maximum height attained by the stone when it is a 2/3 of its total height is calculated as follows;

v² = u² - 2gh

where;

  • v is final velocity at maximum height, v = 0
  • u is initial velocity
  • g is acceleration due to gravity

0 =  u² - 2gh

2gh =  u²

h =  u²/2g

h = (15²)/(2 x 9.8)

h = 11.48 m

h = 11.5 m

Thus, the maximum height that can be reached by the stone is determined as 11.5 m

Learn more about maximum height here: brainly.com/question/12446886

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4 0
2 years ago
A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun
cluponka [151]

Answer:Highest frequency  =618.89Hz

Lowest frequency=582.22Hz

Explanation:

 The linear velocity of a sound generator  is related to angular velocity and is given as

Vs = rω  where

r = the radius of circular path = 1.0 m

ω is the angular velocity of the sound generator. = 100 rpm

1 rev/min = 0.10472 rad/s

100rpm =10.472 rad/ s

Vs = rω

= 1m x 10.472rad/ s=  10.472m/s

A) Highest frequency  heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,

f max = (v/ v-vs) fs

Where , v is the speed of the sound in air at 20 degrees celcius =

343 metres per second

vs is the linear velocity of the sound generator=10.472m/s

fs is the frequency of the sound generator= 600 Hz

f max = (343/ 343 - 10.472) x 600

=343/332.528) x600

=618.89Hz

B) Lowest frequency  heard by a student in the classroom = Minimum frequency

f min = (v/ v+vs) fs

(343/ 343 + 10.472) x 600

=343/353.472) x 600

=582.22hz

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3 years ago
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