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tangare [24]
3 years ago
13

An electric charge of 1 microcoulomb is affected by an electric field of intensity 1000 newtons per coulomb at a point, then thi

s point is a meter away from it by a distance
Physics
1 answer:
Alchen [17]3 years ago
8 0

Answer:

r = 3 m

Explanation:

Given that,

Charge, q=1\ mu C

Electric field at a point, E=1000\ N/C

We need to find the distance away from it. We know that electric field at point r is given by :

E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 10^{-6}}{1000}} \\\\r=3\ m

So, the required distance is 3 m.

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Can anyone please explain this point with an example. I have presentation tomorrow.
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Explanation:

Efficiency is a way of describing the amount of useful ​output​ a process or machine can generate as a percentage of the ​input​ required to make it go. In other words, it compares how much energy is used to do work versus how much is lost or wasted to the environment. The more efficient the machine, the less energy wasted.

For example, if a heat engine is able to turn 75 percent of the fuel it receives into motion, while 25 percent is lost as heat in the process, it would be 75 percent efficient. Out of the original 100 percent of the fuel, 75 percent was output as useful work.  

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8 0
2 years ago
Is direction the length of the route between two points ?
zepelin [54]

Answer:

No distance is the length between two routes

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6 0
2 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
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