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aleksklad [387]
3 years ago
12

Two positive charges of 20 micro coulomb and 100 micro coulomb and the distance between them is 150cm.What will be the electrica

l force of repulsion between these charges?
Physics
1 answer:
Roman55 [17]3 years ago
3 0

Answer:

0.8 N

Explanation:

From coulomb's law,

Formula:

F = kqq'/r²........................ Equation 1

Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.

Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.

Constant:  k = 9×10⁹ Nm²/C²

Substitute these values into equation 1

F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²

F = 1800×10⁻³/2.25

F = 1.8/2.25

F = 0.8 N

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Answer:

37.7 J

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6 0
3 years ago
A single coil of wire, with a radius of 0.13 m is rotated in a uniform magnetic field such that the angle between the field vect
Vedmedyk [2.9K]

The magnitude of emf induced in the single coil of wire rotated in the uniform magnetic field is 0.171 V.


<h3>Induced emf</h3>

The emf induced in the loop is determined by applying Faraday's law of electromagnetic induction.

emf = N(dФ/dt)

where;

  • N is number of turns of the wire
  • Ф is magnetic flux

Ф = BA

where;

  • B is magnetic field strength
  • A is the area of the loop

emf = NBA/t

A = πr²

A = π x (0.13)²

A = 0.053 m²

emf = NBA/t

emf = (1 x 3.746 x 0.053)/(1.16)

emf = 0.171 V

Thus, the magnitude of emf induced in the single coil of wire is 0.171 V.

Learn more about induced emf here: brainly.com/question/13744192

5 0
3 years ago
A source emits sound with a frequency of 16246 Hz. Both it and an observer are moving toward each other, each with a speed of 14
nasty-shy [4]

Answer:

39267.96 Hz

Explanation:

Given:

Frequency emitted = 16246 Hz

Speed of source = 141 m/s

Speed of observer = 141 m/s

Let the velocity of sound is v = 340 m/s

Doppler effect is nothing but the change in the wavelength or frequency of a wave relative to the observer, who is also moving relative to the origin of the wave source.

Therefore, Doppler effect when the source and the observer moves towards each other,

Frequency heard = ( frequency emitted ) ( Velocity of sound + Speed of the observer ) / ( velocity of sound - speed of source )

                      16246 x ( 340 + 141 ) / ( 340 - 141 )

                         = 39267.96 Hz

7 0
3 years ago
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

8 0
3 years ago
Solve for M₂
soldi70 [24.7K]

Explanation:

M₂ = Fr²/GM₁

M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]

M₂ = (7.79N*m²)/(7.84*10^-7N*m²)

M₂ = 9.94*10^6 kg

5 0
3 years ago
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