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Naddik [55]
3 years ago
10

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

Physics
1 answer:
zlopas [31]3 years ago
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<span>.

So,

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

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

</span><span>
</span>
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What kind of energy change occurs when a battery is operating a remote control toy?
solniwko [45]

Answer:

Explanation:

The correct answer is option C.

When the battery is operating a remote control toy the energy is converted from potential energy to the kinetic energy.

A battery stores electrical potential from the chemical reaction.

When the circuit is connected to the potential energy of the battery helps in the movement of the toy.

The energy produced by the movement of the control toy is kinetic energy.

Hence, we can say that Potential energy is changed to kinetic energy

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3 years ago
Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud
Levart [38]

Explanation:

Given that,

Charge 1, q_1=-5.45\times 10^{-6}\ C

Charge 2, q_2=4.39\times 10^{-6}\ C

Distance between charges, r = 0.0209 m

1. The electric force is given by :

F=k\dfrac{q_1q_2}{r^2}

F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}

F = -492.95 N

2. Distance between two identical charges, r=0.0209\ m

Electric force is given by :

F=\dfrac{kq_3^2}{r^2}

q_3=\sqrt{\dfrac{Fr^2}{k}}

q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}

q_3=4.89\times 10^{-6}\ C

Hence, this is the required solution.

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Does the tide that the moon raises on the earth different?
alukav5142 [94]

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Explanation:

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2 years ago
The periodic wave in the diagram below has a frequency of 80. hertz.
ASHA 777 [7]

Answer:

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2 years ago
A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The
cluponka [151]

Answer:

v = 4.1 m/s

Explanation:

As per mechanical energy conservation law we can say that initial total gravitational potential energy of the sphere is equal to final kinetic energy of rolling

so we will say

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

now for pure rolling condition we know that

v = R\omega

so we will have

mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})

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now we will have

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v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}

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