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gogolik [260]
3 years ago
6

For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency

(6.8×109 Hz ) and wavelength (0.044 m ). Recall that the Avogadro constant is 6.022×1023 mol−1. Express the energy in joules to two significant figures.
Physics
1 answer:
Angelina_Jolie [31]3 years ago
3 0

Answer:

2.7 J

Explanation:

The energy of one photon is given by

E=hf

where

h is the Planck constant

f is the frequency

For the photons in this problem,

f=6.8\cdot 10^9 Hz

So the energy of one photon is

E_1=(6.63\cdot 10^{-34})(6.8\cdot 10^9 )=4.5\cdot 10^{-24} J

The number of photons contained in 1.0 mol is

N_A = 6.022\cdot 10^{23} mol^{-1} (Avogadro number)

So the total energy of N_A photons contained in 1.0 mol is

E=N_A E_1 =(6.022\cdot 10^{23})(4.5\cdot 10^{-24})=2.7 J

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Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

<h3>What is the average speed of the train?</h3>

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

  • Distance traveled = 221miles
  • Elapsed time = 3 hours and 40 minutes

First we convert miles to meters and Hours minutes to seconds.

221 miles = ( 221 × 1609.344 )m = 355665.024 meters

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Speed = Distance ÷ time

Speed = 355665.024m / 13200s

Speed = 26.944m/s

Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

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1 year ago
Which car has the most kinetic energy ??
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A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c
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Answer:

Velocity of the car at the bottom of the slope: approximately 20.3\; \rm m \cdot s^{-2}.

It would take approximately 3.9\; \rm s for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

  • Let v denote the final velocity of the car.
  • Let u denote the initial velocity of the car.
  • Let a denote the acceleration of the car.
  • Let x denote the distance that this car travelled.

v^2 - u^2 = 2\, a\cdot x.

Given:

  • u = 3\; \rm m \cdot s^{-1}.
  • a = 4.5\; \rm m \cdot s^{-2}.
  • x = 45\; \rm m.

Rearrange the equation v^2 - u^2 = 2\, a\cdot x and solve for v:

\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}.

Calculate the time required for reaching this speed from u = 3\; \rm m \cdot s^{-1} at a = 4.5\; \rm m \cdot s^{-2}:

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