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Bas_tet [7]
3 years ago
13

Consider a spherical Gaussian surface of radius R centered at the origin. A charge Q is placed. inside the sphere. Where should

the charge be located to maximize the magnitude of the flux of the electric field through the Gaussian surface?
Physics
1 answer:
Elanso [62]3 years ago
5 0
I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. "<span>The charge can be located anywhere, since flux does not depend on the position of the charge as long as it is inside the sphere" gives the answer to the question. I hope that this answer has come to your help.</span>
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Answer:

I am pretty sure it is B.

Explanation:

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Which air mass is responsible for bringing warm, moist air to an area?
Vika [28.1K]

Answer:

Pretty sure it is air mass C

Explanation:

It looks like yhe air has warm air where as the other 2 options have cold air.

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3 years ago
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A nonmechanical water meter could utilize the Hall effect by applying a magnetic field across a metal pipe and measuring the Hal
Serga [27]

Answer:

The velocity is  v =2.455 \  m/s

Explanation:

From the question we are told that  

   The diameter of the pipe is  d =  4.25 \ cm  = 0.0425 \ m

    The magnetic field is  B  =  0.575 \  T

     The hall voltage is  V_H =  60.0 mV  =  60 *10^{-3} \  V

Generally the average fluid velocity is mathematically represented as

          v = \frac{V}{ B  *  d }

=>       v = \frac{60*10^{-3}}{ 0.575   *  0.0425  }

=>       v =2.455 \  m/s

5 0
3 years ago
The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is
Bumek [7]

Answer:

Answer is 25 kg m/s.

Explanation:

Given Data ;

mass = 0.5 kg

velocity = 10 m/s

Find ;

K.E = ?

Formula ;

KE = 1/2 mv 2

Solution:

KE = 1/2(0.5)(10)²(kg)(m/s)

     =25 kg m/s

7 0
3 years ago
Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen
bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

f_{a} =(\frac{v+v_{c} }{v} )f_{s} (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s} (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s

b) Substitute eq. 1 in eq. 2:

f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz

7 0
4 years ago
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