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Oxana [17]
1 year ago
5

A laser emits light with a frequency of 5.69 x 1014 s-1. the energy of one photon of the radiation from this laser is:_________

Physics
1 answer:
Igoryamba1 year ago
8 0

The radiation from this laser is 3.7 x 10^(-19)

<h3>Calculation</h3>

here ∨ = 5.69 x 10^(14) s-1

we know E=h∨

where h = 6.626 x 10^(-34) js

putting the value in the formula, we have,

6.626 x 10^(-34)  x 5.69 x 10^(-34)

=3.7 x 10^(-19)

<h3>What is radiation?</h3>

The energy emission in the form of electromagnetic waves or in the form of moving subatomic particles which are especially high-energy particles which cause ionization is called as the phenomenon of radiation.

Any kind of energy that emanates from a source and moves through space at the speed of light is referred to as radiation. This energy has wave-like qualities and is accompanied by an electric field and a magnetic field.

To know more about radiation, visit:

brainly.com/question/20604369

#SPJ4

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A meteoroid is traveling east through the atmosphere at 18. 3 km/s while descending at a rate of 11.5 km/s. What is its speed, i
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Answer:

The speed of meteoroid is 21.61 km/s in south-east.

Explanation:

Given that,

A meteoroid is traveling through the atmosphere at 18.3 km/s. while descending at a rate of 11.5 km/s it means 11.5 km/s in south.

We need to draw a diagram

Using Pythagorean theorem

AC^2=AB^2+BC^2

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6 0
3 years ago
a crane lifts four pallets of bricks each of which weigh 5000 N. the crane lifts each pallet a height of 30m. the crane takes 4
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Answer:

625 W

Explanation:

Applying

P = W/t.................... Equation 1

Where p = power, W = Work, t = time

But,

W = Force (F) × distance (d)

W = Fd........................ Equation 2

Substitute equation 2 into equation 1

P = Fd/t.................... Equation 3

From the question,

Given: F = 5000 N, d = 30 m, t = 4 munites = (4×60) seconds = 240 seconds

Substitute these values into equation 3

P = (5000×30)/240

P = 625 Watt

3 0
3 years ago
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Explanation:

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3 0
3 years ago
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
adell [148]

Answer:

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

how much work is done on the monitor by (a) friction, (b) gravity

work(friction) = 453.5J

work(gravity) = -453.5J

Explanation:

Given that,

mass = 14kg

displacement length = 5.50m

displacement angle = 36.9°

velocity = 2.30cm/s

F = ma

work(friction) = mgsinθ .displacement

                      = (14) (9.81) (5.5sin36.9°)

                       = 453.5J

work(gravity)

= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)

= 126.9°

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