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AfilCa [17]
2 years ago
11

A source charge generates an electric field of 4286 N/C at a distance of 2. 5 m. What is the magnitude of the source charge? (Us

e k=) 1. 2 µC 3. 0 µC 1. 2 C 3. 0 C.
Physics
1 answer:
svp [43]2 years ago
8 0

The magnitude of the source charge is 3 μC which generates 4286 N/C of the electric field. Option B is correct.

What does Gauss Law state?

It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.

Q = \dfrac {ER^2}k

Where,

E = electric force = 4286 N/C

k = Coulomb constant = 8.99 \times  10^9 \rm\ N m ^2 /C ^2

Q\\
     = charges = ?

r = distance of separation = 2.5 m

Put the values in the formula,

Q  = \dfrac {4286\times  2.5 ^2}{8.99 \times  10^9 }\\\\
Q  = 3\rm \  \mu C

Therefore, the magnitude of the source charge is 3 μC.

Learn more about Gauss's law:

brainly.com/question/1249602

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<h3>What is density?</h3>

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What is input and output in science physics​
tresset_1 [31]

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Input work is the work done on a machine as the input force acts through the input distance .Other ways it would be the work done on a body or system, that is, forces that are applied to the body or system. This is in contrast to output work which is a force that is applied by the body or system to something else. Output work is the work done by a machine as the output force acts through the output distance. The machine does to the object to increase the output distance.

Explanation:

8 0
3 years ago
Consider a system consisting of three particles: m1 = 4 kg, vector v1 = &lt; 10, -5, 15 &gt; m/s, m2 = 9 kg, vector v2 = &lt; -1
Ivahew [28]

The answers to your questions are as listed below

A) The total momentum of the system is :  < - 167, 133, 87 >   kg m/s

B) Velocity of the center of mass of this system : < - 10.44 , 8.31, 5.44 > m/s

C) Total kinetic energy of the system : 5159.5 J  

D) Translational kinetic energy of the system :  1660.84 J

E) kinetic energy relative to the center of mass : 3498.66 J

<u>Given data : </u>

m₁ = 4 kg ,   V₁ = < 10, -5, 15 > m/s

m₂ = 9 kg,   V₂ = < -15, 5, -3 > m/s

m₃ = 3 kg    V₃ = < -24, 36, 18 > m/s.

<h3>A) Determine the total momentum of the system </h3>

Total momentum = Σ mv

= < 4*10 + 9*(-15) + 3*(-24), 4*(-5) + 9*5 + 3*36, 4*15 + 9 (-3) + 3*18 >

= < - 167, 133, 87 >   kg m/s

<h3> B) Velocity at the center of mass of this system </h3>

Velocity = < - 167 / 16 , 133/ 16, 87/ 16 >

              =  < - 10.44 , 8.31, 5.44 > m/s

<h3>C) Determine the total kinetic energy of the system </h3>

Total Kinetic energy = 1/2 m₁v₁² + 1/2 m₂v₂²  + 1/2 m₃v₃²  --- ( 1 )

where : v² = v₁² + v₂² + v₃²

therefore

Total kinetic energy = 700 + 1165.5 + 3294

                                  = 5159.5 J  

<h3>D) Determine the translational kinetic energy of the system </h3>

Translational kinetic energy ( Kcm ) = 1/2 ( m₁ + m₂ + m₃ ) * Velocity²

                                                           = 1/2 ( 16 ) * (  - 10.44² + 8.31² + 5.44² )

Hence translational kinetic energy = 1660.84 J

E) The Kinetic energy of the system relative to the center of mass

Krel = Ktotal - Kcm

       = 5159.5 - 1660.84

       = 3498.66 J

Hence we can conclude that the answers to your questions are as listed above.

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