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ANEK [815]
3 years ago
8

Technician A says if a fuse burns out when a switch is turned on, a short circuit is present before the switch. Technician B say

s if a fuse burns out when a switch is turned on, a short circuit is present after the switch. Who is correct?
Physics
1 answer:
lord [1]3 years ago
7 0

Answer:

Technician B

Explanation:

This can be explained as a fuse is a protective device which blows in order to protect the circuit and the equipment from short circuiting.

Short circuit is a condition where a very heavy current flow through the line with no low line impedance which is not the intended path for the current to flow.

Thus when heavy current flows through a fuse it blows out thus disconnecting the path for the current flow.

Therefore, if a fuse burns out when switch is on, short circuit is present after the switch

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The figure shows five electric charges. Four charges with the magnitude of the charge 2.0 nC form a square with the size a = 4.0
bearhunter [10]

Considering the four electric charges forming a square with magnitude of charge of 2.0 nC on each. :

A) Magnitude of the force on the 5.5 nC charge in the middle of the figure = 3.48 * 10^-4 N

B) Direction of the force on the 5.5 nC charge in the middle of the figure = Leftward ( negative x -axis )

Using the given data :

size of square = 4 cm

magnitudes of four charges = 2.0 nC

<u> a) </u><u>magnitude</u><u> of the force on the center charge </u>

Electric force between two point charges = F = \frac{1}{4\pi *E_{o} } \frac{q1q2}{r^2} ----- ( 1 )

where ; \frac{1}{4\pi E_{o} } = 9 * 10^9 Nm^2/C^2

step 1 ; find r ( distance between charges )

r² = ( 2 )² + ( 2 )² = 8 cm²  

back to equation 1

F = 9 * 10⁹ * \frac{2 * 10^{-9} *  (5.5 * 10^{-9})  }{8*10^{-4} } =  1.23 * 10^-4  N

∴ magnitude of the force on the center charge ( Fnet )= 4F cos 45°

        = 4 * ( 1.23 * 10^-4 ) * \frac{1}{\sqrt{2} }  = 3.48 * 10^-4 N

b) The direction of the force at the center is along the negative x-axis ( leftward )

Learn more : brainly.com/question/24139734

6 0
3 years ago
Consider two identical objects released from rest high above the surface of Earth. (Neglect air resistance for this question.)
Nadusha1986 [10]

The question is missing its alternatives. Here is the complete question.

Consider two identical objects released from rest high above the surface of Earth. In Case 1, the object is released from a height above the surface of Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1. In Case 2, the obejct is released from a height above the surface of Earth equal to 2 Earth radii and its kinetic energy just before it hits is K2.

1. Compare the kinetic energy of the two objects just before they hit the surface of the earth.

a) K2 = 2K1; b) K2 = 4K; c) K2 = (4/3)K1; d) K2 = (3/2)K1;

Answer: C) K2 = (4/3)K1

Explanation: As it is related to the gravity of the Earth, the potencial energy is: U(r)= - \frac{G.Me.m}{r} + U₀

In this case, U₀=0, G is the universal gravitational constant, Me is the mass of Earth, m is the mass of the object and r is the distance between the center of the Earth and the object.

The potencial energy of an object of mass m on the surface of the Earth is:

Usurface = - \frac{G.Me.m}{Re}

The potencial energy of the object in Case 1 is

U1 = - \frac{G.Me.m}{2Re}

For the Case 2:

U2 = - \frac{G.Me.m}{3Re}

The potencial change in Case 1:

ΔU1 = - G.Me.m.(\frac{1}{Re}-\frac{1}{2Re}) = - \frac{1}{2}\frac{G.Me.m}{Re}

For Case 2:

ΔU2 = - G.Me.m(\frac{1}{Re}-\frac{1}{3Re}) = - \frac{2}{3}\frac{G.Me.m}{Re}

Comparing ΔK1 and ΔK2 equals comparing ΔU1 and ΔU2:

Δ\frac{U2}{U1} = (-2/3)(-1/2) = 4/3

So, comparing kinetic energies, K2 is 4/3 of K1.

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The waves lose energy in the form of heat. The frequency of the waves continues to increase. Disturbances inside Earth's core cause earthquakes. The starting point of the disturbance is called the epicenter.

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Hitman42 [59]

Answer:

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Answer:

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