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blondinia [14]
4 years ago
15

Light with a wavelength of 5.0 · 10-7 m strikes a surface that requires 2.0 ev to eject an electron. Calculate the energy, in jo

ules, of one incident photon at this frequency.
_____ joules
Physics
2 answers:
GenaCL600 [577]4 years ago
7 0

6.2 x 10^-13 is your answer

Aleonysh [2.5K]4 years ago
7 0

Answer:The energy, in joules, of one incident photon at this frequency3.97\times 10^{-19} Joules.

Explanation:

The wavelength of the light = \lambda =5.0\times 10^{-7} m

Energy of a single photon is given by the expresssion:

E=\frac{hc}{\lambda }

h = Planck's constant

c = speed of light

E=\frac{6.626\times 10^{-34} J.s\times 3\times 10^8 m/s}{5.0\times 10^{-7} m}=3.97\times 10^{-19} Joules

The energy, in joules, of one incident photon at this frequency3.97\times 10^{-19} Joules.

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What are the similarities between strong nuclear force and weak nuclear force
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3 years ago
A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
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stiks02 [169]

Answer:

1 / 2 m v^2 = L m g (1 - cos θ)

This is the KE due to the pendulum falling from a 25 deg displacement

v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2

v = 1.92 m/s      this is the speed due to an initial displacement of 25 deg

Its speed at the bottom would then be

1.92 + 1.2 = 3.12 m/s   since it gains 1.92 m/s from its initial displacement

3 0
1 year ago
What is the main reason why many nuclear power plants are located near bodies of water? to wash wastes out of the power plant to
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Answer: to avoid problems with water supply

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Read 2 more answers
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