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Maurinko [17]
3 years ago
12

A circuit breaker is rated for a current of 25 a rms at a voltage of 240 v rms. (a) what is the largest value of imax that the b

reaker can carry?
Physics
1 answer:
Mademuasel [1]3 years ago
8 0

The largest value of current that the breaker can carry = Imax=35.4A

Explanation:

Rms value of current= Irms= 25 A

The rms current and the maximum current are related as

Imax= √2 Irms

Imax=√2 (25)

Imax=35.4 A

Thus the maximum current carried by the breaker= 35.4 A

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Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

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Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

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differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

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a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

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Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

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Part c)

For velocity of the particle we have

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now at t = 1.1 s

\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

\theta = -127.7 degree

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