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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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kramer

Answer:

8.08 J/g °C

Explanation:

Q=m*Cp*ΔT-->

Cp=Q/(m*ΔT) -->

Cp=1600/[18*(31-20)]-->

Cp=8.08 J/g °C

6 0
3 years ago
A.) If its booster rockets accelerate the space shuttle at 15m/s2, how high will it be one minute after launch?
poizon [28]

Answer:

27,000 m

450 m/s

Explanation:

Assuming the initial velocity is 0 m/s:

v₀ = 0 m/s

a = 15 m/s²

t = 60 s

A) Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²

Δy = 27,000 m

B) Find: v_avg

v_avg = Δy / t

v_avg = 27,000 m / 60 s

v_avg = 450 m/s

5 0
3 years ago
Ignoring friction, what is the acceleration of a 75.0 kg object which is acted upon by an
natulia [17]

Explanation:

we can use the formula F = ma

hence,

500 = 75a

a = 6.666 m/s² or 6(2/3) m/s²

hope this helps.

3 0
2 years ago
The distance between two parallel wires carrying currents of 10 A and 20 A is 10 cm. Determine the magnitude and direction of th
Marianna [84]

Explanation:

It is given that,

Current in wire 1, I₁ = 10 A

Current in wire 2, I₂ = 20 A

Distance between wires, d = 10 cm = 0.1 m

Force per unit length is given by :

\dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}

\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 10\times 20}{2\pi \times 0.1}

\dfrac{F}{l}=0.0004\ N/m

\dfrac{F}{l}=4\times 10^{-4}\ N/m

So, the magnetic force acting per unit length of the wires 4\times 10^{-4}\ N/m. Since, the current is in same direction. So, the force is attractive in nature.

6 0
3 years ago
The kinetic energy, k, of an object varies jointly with the mass of the object and the square of the velocity of the object. a 5
o-na [289]

Answer:

KE = 1/2 M V^2 = 1/2 * 25 * 10^2 = 1250 J

Check

M2 = 1/2 M1

V2 = V1 / 2

E2 = 1/2 * 1/4 E1 = E1 / 8 = 10000 / 8 = 1250 J

8 0
2 years ago
Read 2 more answers
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