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3 years ago

Coulomb is a very large unit for practical use. Justify your answer if 10^10 electrons are transferred from a body/second

1 answer:

3 0

Given:

10^10 electrons per second

To justify that coulomb is a very large unit for practical use, we need to convert the quantity of electron given to Coulombs:

From literature,

1 Coulomb is equivalent to 6.242×10^18 electrons.

So,

= 10^10 electrons * (1 coulomb/6.242×10^18 electrons) / second
= 1.602 x 10^-9 coulumbs

This value is too small to be used in an actual setting.

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3 years ago

A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the

blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

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3 years ago

A car takes off from rest takes of from rest and covers a distance of 80m on a straight road in 10s.Calculate the magnitude of i

hodyreva [135]

Initial velocity (u) = 0 m/s [the car was at rest]
Distance (s) = 80 m
Time (t) = 10 s
Let the magnitude of acceleration be a.
By using the equation of motion, $s = ut + \frac{1}{2} a {t}^{2}$ we get, $80 = 0 \times 10 + \frac{1}{2} \times a \times {10}^{2} \\ = > 80 = \frac{1}{2} \times 100a \\ = > 80 = 50a \\ = > a = \frac{80}{50} \\ = > a = 1.6$

Answer:

The magnitude of its acceleration is 1.6 m/s^2.

Hope you could get an idea from here.

Doubt clarification - use comment section.

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3 years ago