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3 years ago

What is the average electric current in a wire, when a charge of 150 C passes in 30 s?

1 answer:

7 0

We know, I = Q / t
Here, Q = 150 C
t = 30 s

Substitute their values,
I = 150 / 30
I = 5 A

In short, Your Answer would be 5 Ampere

Hope this helps!

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An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

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14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

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3 years ago

A guitar string is 0.620m long, and oscillates at 234Hz. What is the velocity of the waves in the string? m/s

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Answer:

v = 72.54 m/s

Explanation:

We have,

Length of a guitar string is 0.62 m

Frequency of a guitar string is 234 Hz

For guitar string,

$L=2\lambda\\\\\lambda=\dfrac{L}{2}\\\\\lambda=\dfrac{0.62}{2}\\\\\lambda=0.31\ m$

The velocity of the wave in the string is given by :

$v=f\lambda\\\\v=234\times 0.31\\\\v=72.54\ m/s$

So, the velocity of the waves in the string is 72.54 m/s.

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3 years ago

Explain the origin of the magnitude designation for determining the brightness of stars. Why does it seem to go backward, with s

Mashcka [7]

Answer:

Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.

This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.

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3 years ago