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aniked [119]
3 years ago
5

The unit used to measure electric current is the ampere (A). Now, assume that the current delivered at a wall socket reaches the

value 3.8 A ten times in a time interval of 0.17 s. What is the period T with which the current at the wall socket changes?
Physics
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

T = 0.017s

Explanation:

period is the time it takes a particle to make one oscillation

An electric current is periodic in nature

The current reaches 3.8A  ten times.  

So there must have been 10 cycles (10 periods) in 0.17s.   let 'T'  be the period:

T=\frac{t}{n}

t is the total time interval

n is the number of oscillations

T=\frac{0.17}{10}

10T = 0.17

T = 0.17/10 = 0.017s

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Answer:

F=5.8\times 10^{8}\ N

F=35.57\times 10^{21}\ N

Explanation:

Given that

Intensity I

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As we know that pressure due to intensity given as

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P=4.6\times 10^{-6}\ Pa

We know that force F

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F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N

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b)Gravitational force F

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M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg

r =1.5\times 10^{11}\ m

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So F

F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}

F=35.57\times 10^{21}\ N

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