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3 years ago

Dimension for electric charge

Physics

1 answer:

Artemon [7]3 years ago

7 0

Answer:

Explanation:

Dimensions of current = A

Dimensions of time = T

Current = Charge / time

Therefore Charge = Current × time

[Charge] = [Current] × [time]

= AT

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when an object moves with constant velocity, does its average velocity during any time interval differ from its instataneous vel

BigorU [14]

For a constant-velocity object, the average and instantaneous are the same. So the answer is no. It's like taking a running average of a string of numbers that are all the same number. The average is always the sum of the numbers divided by how many have accumulated, which will always equate to the repeated number.

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3 years ago

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

notsponge [240]

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

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

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

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

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3 years ago

Increasing which two things would increase gravitational potential energy?

sukhopar [10]

Answer:

Mass and height

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about $9.8 m/s2.$

Which is represented as;

$PE_g=mgh$

$PE_g$ stands for gravitational potantial energy,

m stands for mass of object,

g is the gravitational constant and

h is the height.

Here we see that mass of object and height is directly proportional to the gravitational potential energy.

That means increasing in mass and height will result in increasing gravitational potential energy.

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3 years ago

Li and Raj are playing football. They take turns trying to run past each other with the football. Li weighs 125 pound and Raj we

Elenna [48]

Answer:

Li has less mass and therefore less inertia, so he can change his motion more easily than Raj.

Explanation:

Inertia describes the resistance of an object to any change in its state of motion, and it depends on the mass of the object only. In particular:

- if an object has a large inertia (large mass), then it is more difficult to change its state of motion (i.e. to put it in motion, or to slow it down, or to change its direction of motion)

- if an object has small inertia (small mass), then it is more easy to change its state of motion

In this problem, Li has less mass than Raj, so he has less inertia, therefore he can change his motion more easily than Raj.

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3 years ago

Read 2 more answers

Define potential difference in a uniform electric field?

8090 [49]

Explanation:

For example, a uniform electric field E is produced by placing a potential difference (or voltage) ΔV across two parallel metal plates, labeled A and B. ... The relationship between ΔV and E is revealed by calculating the work done by the force in moving a charge from point A to point B.

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2 years ago