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gogolik [260]
3 years ago
7

A carbon dioxide laser produces radiation with a wavelength of 1.06 x 104 nm. What is the frequency of this radiation?

Physics
1 answer:
olga55 [171]3 years ago
3 0

Answer:

frequency of the radiation will be 2.83\times 10^4Hz

Explanation:

We have given wavelength of the radiation is 1.06\times 10^4m

Velocity of the light c=3\times 10^{8}m/sec

We have to find the frequency of the radiation

We know that velocity of light is given by v=\lambda f, here \lambda is wavelength and f is frequency

So frequency of radiation will be f =\frac{v}{\lambda }=\frac{3\times 10^8}{1.06\times 10^4}=2.83\times 10^4Hz

So frequency of the radiation will be 2.83\times 10^4Hz

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3 years ago
A machine does 1500 joules of work in 30 second What is the power of the machine
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8 0
3 years ago
Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp
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Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the  final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M  = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

8 0
3 years ago
A coil of wire with 100. circular turns of radius 8.00 cm is in a uniform magnetic field along the axis of the coil. The resista
andreyandreev [35.5K]
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6 0
3 years ago
A boat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, space, start fraction, m, divided by, s,
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Answer:

d=42m(-\hat{i})

Explanation:

For this problem, we need to apply the formulas of constant accelerated motion.

To obtain the boat displacement we need to calculate the displacement because of the river flow and the displacement done because of the boat motor.

for the river:

d_r=v*t\\d_r=5m/s*6s\\d_r=30m(\hat{i})

for the boat:

x=\frac{1}{2}*a*t^2\\\\x=\frac{1}{2}*4.0m/s*(6s)^2\\\\\\x=72m(-\hat{i})

So the final displacement is given by:

d=dr+x\\d=30m-72m\\d=42m(-\hat{i})

8 0
3 years ago
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