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Svetlanka [38]
3 years ago
13

Please helppppppp!!!!!!!!!!!!!!

Physics
1 answer:
azamat3 years ago
8 0

Answer:

circuit breaker

Explanation:

A circuit breaker is a device used for electrical safety. It consists of a switch designed to protect an electrical circuit from damage that may result from heating due to overload in the circuit.

Its basic function is to interrupt current flow through its switch that consists of metal stripe which bends when it gets hot.

Fuse has similar action with circuit breaker, the only difference is that fuse can only be used once because it melts when it gets hot.

Therefore, the correct answer is "circuit breaker"

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<u>The simple method:
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E=hf

therefore f=e/h

f=(3.611x10^-15) / 6.63x10^-34)

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An object accelerates from rest at a constant rate covering a distance of 18 meters in 3 seconds. What is its rate of accelerati
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4

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Describe the motion of an object that has an acceleration of 0m/s2
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Tungsten has a temperature coefficient of resistivity of 0.0045 (c°)-1. a tungsten wire is connected to a source of constant vol
Murljashka [212]

Answer:

131.1^{\circ}C

Explanation:

The power delivered in the wire is given by:

P=\frac{V^2}{R}

where V is the voltage of the battery and R is the resistance of the wire.

Since the voltage of the battery is constant, we can rewrite this equation as follows:

V^2 = PR=const. (1)

At the beginning, the initial resistance is R_0, and the power delivered is P_0. Later, when the temperature increases, the power becomes P_1 = \frac{2}{3}P_0, and the new resistance is R_1. Using (1), we can write

P_0 R_0 = \frac{2}{3}P_0 R_1\\R_1 = \frac{3}{2}\frac{P_0 R_0}{P_0}=\frac{3}{2}R_0 (2)

So, the new resistance must be 3/2 of the initial resistance.

We know that the resistance increases linearly with the temperature, as

R_1 = R_0 (1+\alpha \Delta T)

where

\alpha = 0.0045 ^{\circ}C^{-1} is the temperature coefficient

\Delta T is the change in temperature

Using (2), we can rewrite this equation as

\frac{3}{2}R_0 = R_0(1+ \alpha \Delta T)

and we find:

\frac{3}{2}=1+\alpha \Delta T\\\Delta T=\frac{\frac{3}{2}-1}{\alpha}=111.1 ^{\circ}

So, the new temperature of the wire must be

T_f = 21^{\circ}+111.1^{\circ}=132.1^{\circ}

6 0
3 years ago
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