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sergey [27]
4 years ago
5

Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through polystyrene. The waveleng

th of the light from the laser is 632.8 nm in air and the index of refraction of polystyrene is 1.490.
Physics
1 answer:
klemol [59]4 years ago
3 0

Answer:

Speed:

2.01x10^{8}m/s

Wavelength:

4.24x10^{-7}m

Frequency:

4.74x10^{14}Hz

Explanation:

The speed of the laser as it travels through polystyrene can be determine by means of the equation of the refraction index:

n = \frac{c}{v} (1)

Where c is the speed of light and v is the speed of the laser in the medium.

Therefore, v will be isolated from equation 1

v = \frac{c}{n}

v = \frac{3x10^{8}m/s}{1.490}

v = 2.01x10^{8}m/s

Hence, the speed of the laser has a value of 2.01x10^{8}m/s

Frenquency:

Since, wavelength is the only one who depends on the media. Therefore the frequency in both medium will be the same.  

To determine the frequency it can be used the following equation

c = \nu \cdot \lambda  (2)

Where c is the speed of light, \nu is the frequency and \lambda is the wavelength

Then, \nu wil be isolated from equation 2.

\nu = \frac{c}{\lambda}  (3)

Before using equation 3 it is necessary to express \lamba in units of meters.

\lambda = 632.8nm . \frac{1m}{1x10^{9}nm} ⇒ 6.328x10^{-7}m

\nu = \frac{3x10^{8}m/s}{6.328x10^{-7}m}

\nu = 4.74x10^{14}s^{-1}

\nu = 4.74x10^{14}Hz

Hence, the frequency of the laser has a value of 4.74x10^{14}Hz

Wavelength:

To determine the wavelength it can be used:

v = \nu \cdot \lambda

\lambda = \frac{v}{\nu}

Where v is the speed of the laser through the polystyrene.

\lambda = \frac{2.01x10^{8}m/s}{4.74x10^{14}s^{-1}}

\lambda = 4.24x10^{-7}m

Hence, the wavelength of the laser has a value of 4.24x10^{-7}m

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How much force is needed to slow down a 15 kg car traveling at 60 m/s to 15 m/s in 10 seconds?​
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Answer:

The force needed to slow down the car is, F = 67.5 N

Explanation:

Given data,

The mass of the car, m = 15 kg

The initial velocity of the car, V = 60 m/s

The final velocity of the car, v = 15 m/s

The time period of deceleration, t = 10 s

The difference in the momentum of the car is,

                                     mV - mv = 15(60 - 15)

                                                    = 675 kg m/s

The rate of change in momentum of the car gives the force acting on it.

                                    F = (mV - mu) / t

Substituting the values,

                                   F = 675 / 10

                                      = 67.5 N

Hence, the force needed to slow down the car is, F = 67.5 N

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3 years ago
A constant force of 5.00 N acts on a 2.50 kg object for 10.0 s. What are the changes in the object’s momentum and velocity?
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What is the momentum of a vehicle that has a mass of 1500kg and a velocity of 15m/s? (don't forget the unit is kg*m/s, and needs
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Answer:

<h2>22500 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 1500 × 15

We have the final answer as

<h3>22500 kg.m/s</h3>

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3 years ago
How much work would it take to push two protons very slowly from a separation of 2.00×10−10 m (a typical atomic distance) to 3.0
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We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?

The work done is equal to the electric potential energy gained by the proton:

W=q \Delta V

where q=1.6 \cdot 10^{-19}C is the charge of the proton and \Delta V is the potential difference between the final position and the initial position of the proton. To calculate this \Delta V, we must calculate the electric potential generated by the proton at rest at the two points, using the formula:

V=k\frac{Q}{r}

where k=9.0 \cdot 10^9 N m^2 C^{-2} is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find

V_i = (9.0 \cdot 10^9 )\frac{1.6 \cdot 10^{-19}}{2.0 \cdot 10^{-10}}=7.2 V

V_f = (9.0 \cdot 10^9 )\frac{1.6 \cdot 10^{-19}}{3.0 \cdot 10^{-15}}=4.8 \cdot 10^5 V

Therefore, the work done is

W=q \Delta V=(1.6 \cdot 10^{-19}C)(4.8 \cdot 10^5 V-72 V)=7.7 \cdot 10^{-14} J

6 0
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