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Zepler [3.9K]
3 years ago
11

Fred points a toy laser gun at a wall. Considering that the frequency of the gun's light is 4.91 × 1014 hertz and that Planck’s

constant is 6.61 × 10-34 joule·seconds, what is the energy of the laser light?
A) 3.24 × 10-21 joules
B) 4.91 × 1015 joules
C) 2.04 × 10-21 joules
D) 4.92 × 10-18 joules
Physics
2 answers:
vladimir1956 [14]3 years ago
6 0
E=hf 

<span>E=6.61x10^34x4.91x10^14 </span>
<span>E=3.24x10^21 joules </span>

<span>Answer is A </span>
Stella [2.4K]3 years ago
4 0
The answer is 3.24 x 10^-21
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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
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Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

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6 0
3 years ago
Which of the following situations would cause the greatest decrease in the motion of molecules in a system?
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7 0
2 years ago
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2 years ago
A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A,
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Answer:

33,458.71 turns

Explanation:

Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter

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Solution:

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2 years ago
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