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nexus9112 [7]
3 years ago
9

Helpp pls

Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

The intensity of the electric field is

|E|=10654.37 \:N/C

Explanation:

The electric field equation is given by:

|E|=k\frac{q}{d^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge at 0.4100 m from the balloon
  • d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}

|E|=10654.37 \:N/C

I hope it helps you!

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Answer:

the number of photons of yellow light does the lamp generate in 1.0 s is 7 x 10^{19}

Explanation:

given information:

power, P = 25 W

wavelength. λ - 580 nm = 5.80 x 10^{-7} m

time, t = 1 s

to calculate the number of photon(N), we use the following equation

N = λPt/hc

where

λ = wavelength (m)

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t = time interval (s)

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So,

N = λPt/hc

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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.

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