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RUDIKE [14]
3 years ago
13

As it travels through a crystal, a light wave is described by the function E(x,t)=Acos[(1.57×107)x−(2.93×1015)t]. In this expres

sion, x is measured in meters and t is measured in seconds.
Part A
What is the speed of the light wave?
Express your answer to three significant figures and include appropriate units.
Physics
1 answer:
Drupady [299]3 years ago
7 0

Answer:

Speed, v=1.86\times 10^8\ m/s

Explanation:

It is given that,

A light wave is described by the following function as :

E(x,t)=A\ cos[(1.57\times 10^7)x-(2.93\times 10^{15})t].....(1)

The general equation of wave is given by :

E=Acos(kx-\omega t)........(2)

On comparing equation (1) and (2)

k=(1.57\times 10^7)

\dfrac{2\pi}{\lambda}=(1.57\times 10^7)

\lambda=\dfrac{2\pi}{(1.57\times 10^7)}

Wavelength, \lambda=4.002\times 10^{-7}\ m

\omega=(2.93\times 10^{15})

\dfrac{2\pi}{T}=(2.93\times 10^{15})

\dfrac{1}{T}=\dfrac{(2.93\times 10^{15})}{2\pi}

Frequency, f=4.66\times 10^{14}\ Hz

Let v is the speed of the light wave. It is given by :

v=f\times \lambda

v=4.66\times 10^{14}\ Hz\times 4.002\times 10^{-7}\ m

v=1.86\times 10^8\ m/s

So, the speed of the light wave is 1.86\times 10^8\ m/s. Hence, this is the required solution.

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Read 2 more answers
What is its diameter when the temperature is raised to 100 degrees Celsius? (b) What temperature change is required to increase
alexandr1967 [171]

The question is incomplete, the complete question is;

A spherical steel ball bearing has a diameter of 2.540cm at 25.00∘C.

(a) What is its diameter when it's temperature is raised to

100∘C?

(b) What temperature change is required to increase its volume by

1.000% ?

Answer:

a) 2.542 cm

b) 303.03°C

Explanation:

Given;

Diameter of the ball= 2.540cm

Initial temperature= 25.0°C

Final temperature= 100.0°C

Percentage increase in volume = 1.000%

Temperature coefficient of expansion for steel =11.0×10^−6/∘C

d2= d1[1 + α(T2-T1)]

d2= 2.540[1 + 11.0×10^−6(100-25)]

d2= 2.540[1 + 8.25×10^-4]

d2= 2.542 cm

From;

%V ×1/100 = V ×3α ×∆T/ V

Substituting values;

1.000 ×1/100= 3× 11.0×10^−6 × ∆T

∆T= 0.01/3× 11.0×10^−6

∆T= 303.03°C

4 0
3 years ago
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