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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 magnitude of the electrical force acting between a +2.4x10-8c charge and 1+1.8x10-6 charge that are separated by 1.008m is
Temka [501]

Answer:

3.83×10¯⁴ N

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +2.4x10¯⁸ C

Charge 2 (q₂) = +1.8x10¯⁶ C

Distance apart (r) = 1.008 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

The magnitude of the electrical force acting between the two charges can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²

F = 0.0003888 / 1.016064

F = 3.83×10¯⁴ N

Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N

5 0
2 years ago
Assume that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic inte
faltersainse [42]

Answer:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all - option A

Explanation:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all; the reason being that: no magnetic field is being produced by a charge that is static. Only a moving charge can produce a magnetic effect. And the magnet can not have any torque due to its own magnetic lines of force.

5 0
3 years ago
A man has a power of 90 W and mass 60 kg runs up a staircase in 40 s. If each staircase is 20cm high find the number of steps? (
34kurt
I showed my working in the images above. if you have any questions please feel free to ask

4 0
2 years ago
An atom with the expected number of neutrons, protons, and electrons is called a(n)
ryzh [129]

Answer:

Stable atom

Explanation:

A stable atom is one that has a balanced nuclear inter-particle force reaction as such the binding energy of a stable atom is sufficient to permanently keep the nucleus as one unit. Examples of a stable atom are the atoms of  monoisotopic elements such as fluorine, sodium, iodine, gold, aluminium, and cobalt.

In a stable atom the expected number of proton, neutron, and electron are present while in an unstable atom or radioactive atom, there are more than the expected number of neutrons or protons, such that the internal energy of the nucleus is excessive and more than the binding energy, which can lead to radioactive decay.

6 0
3 years ago
2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he
Ratling [72]

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, v_s = d/t₂

∴ v_s = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

5 0
3 years ago
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