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3 years ago

10

Suppose that the voltage is reduced by 10 percent (to 90 VV). By what percentage is the power reduced? Assume that the resistanc

e remains constant.

1 answer:

pickupchik [31]3 years ago

5 0

Answer:

The power is reduced by 19 percent.

Explanation:

The formula of power is given by:

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

In which V is the voltage, and R is the resistance.

I am going to use R = 1 in both cases.

With the original voltage, V = 1, we have

$P = \frac{V^{2}}{R} = \frac{1}{1} = 1$

With the modified voltage, V = 0.9, we have:

$P = \frac{V^{2}}{R} = \frac{0.9^{2}}{1} = 0.81$

So the power is reduced by 1-0.81 = 0.19 = 19 percent.

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3 years ago

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Explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".

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Let X the random variable of interest. We know on this case that $X\sim Exp(\lambda)$

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The seers were actually of the opinion that a healthy mind resides in a healthy body. It implies that there is connection between the body and the mind. When the body catches an illness, the mind and other parts of the body are affected. When our minds are not healthy, it affects the effective functioning of the body.

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2 years ago

Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin

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Now for parallel pipes

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Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

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Thus the answer is option A). 10

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3 years ago