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mart [117]
3 years ago
14

(b) A Blu-ray laser has a power of 5 milliwatts (1 watt = 1 J s−1). How many photons of light are produced by the laser in 1 hou

r? Energy=(Power)x(time) 5x10-3 J s-1x 1 x 3600s=18.0 J
Physics
1 answer:
worty [1.4K]3 years ago
5 0

Answer:

The number of photons of light produced by the laser in 1 hour is

1 Photon / hour

Explanation:

Number of photons of light produced is given by

Number of photons = \frac{Power}{Energy}

From the question,

Power = 5 mW (milliwatts) = 5 × 10⁻³ W

Since 1 Watt = 1 Js⁻¹

Then, 5 × 10⁻³ W = 5 × 10⁻³ Js⁻¹

For the Energy,

As given from the question

Energy=(Power)x(time)

Time = 1 hour = (1 × 60 × 60) s = 3600 s

∴ Energy = 5 × 10⁻³ Js⁻¹ x 1 x 3600s

Energy =18.0 J

Now for the Number of photons produced,

Number of photons = \frac{Power}{Energy}

Power = 5 × 10⁻³ Js⁻¹ = 0.005 Js⁻¹

Number of photons = \frac{0.005}{18}

Number of photons = 2.78 × 10⁻⁴ Photons / sec

This is the number of photons produced in 1 second.

For the number of photons produced in 1 hour, we will multiply the result by 3600

(NOTE: 1 sec = \frac{1}{3600} hour)

Number of photons = 2.78 × 10⁻⁴ Photons / sec

= 2.78 × 10⁻⁴ × 3600 Photons / hour

= 1.0008 Photons / hour

≅ 1 Photon / hour

Hence, the number of photons of light produced by the laser in 1 hour is

1 Photon / hour

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Explain why selling cereal by mass rather then by volume be more fair to customers
Orlov [11]

<em>Answer:</em>

Simple, there could be air in the package and volume would record that, whereas mass would count the mass of the cereal and discount the air.

7 0
3 years ago
An object is 27.0 cm from a concave mirror of focal length 15.0 cm. find the image distance.
pantera1 [17]

The distance of the Image will be -33.75 cm

A concave mirror has an inward-curving reflecting surface that faces away from the light source. Unlike convex mirrors, a concave mirror's image forms a variety of images based on the object's proximity to the mirror.

Given that, an object placed 27 cm from a concave mirror having the focal length of 15 cm

We have to find distance of the Image

Using Mirror Formula:

1/f = 1/v + 1/u

Where,

f = focal length

v =  Image distance from the mirror

u = object distance from the mirror (concave)

Substitute the known values in the above formula to find the value of 'v' i.e. from the mirror.

1/(-15) = 1/v + 1/(-27)

1/(-15) = 1/v - (1/27)

1/v = -0.029

v = -33.75 cm

Therefore the distance of the Image will be -33.75 cm

Learn more about concave mirror here:

brainly.com/question/9816370

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8 0
2 years ago
Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy
Dominik [7]

a) 0.261 T

b) this field strength is obtainable with today's technology

Explanation:

a)

The force experienced by a charged particle moving perpendicular to a magnetic field is given by

F=qvB

where

q is the charge

v is the velocity

B is the strength of the field

This force is perpendicular to the motion of the particle, which therefore moves in a circular path; and so, this force acts as centripetal force, so we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the circle

In this problem, we have:

q=1.6\cdot 10^{-19}C (magnitude of the charge of antiprotons)

v=5.00 \cdot 10^7 m/s (velocity)

m=1.67\cdot 10^{-27}kg (mass of antiprotons)

r = 2.00 m (radius)

Therefore, we can re-arrange the equation and solve to find B, the magnetic field strength:

B=\frac{mv}{qr}=\frac{(1.67\cdot 10^{-27})(5.00\cdot 10^7)}{(1.6\cdot 10^{-19})(2.00)}=0.261 T

B)

The strength of the magnetic field calculated in part A) is

B=0.261 T

This is indeed a very strong magnetic field. In fact, by comparison, the Earth's magnetic field has a strength of about

B_{earth}=5\cdot 10^{-5} T

However, there are current technologies available that are able to produce such strong fields. For instance, the superconducting magnets in the LHC (Large Hadron Collider) are able to produce magnetic fields of strength up to 8 Tesla (8 T).

Therefore, we can say that this field strength is obtainable with today's technology.

5 0
3 years ago
pakka acquires a velocity of 72 km per hour in 10 seconds starting from rest find a ) the acceleration b)the average velocity c)
nadezda [96]
Acceleration equals 24 km/s
Average equals 396km/s
8 0
4 years ago
If the magnetic field in a traveling EM wave has a peak value of 17.9nT, what is the peak value of the electric field strength?
Doss [256]

Answer:

5.37 N/C

Explanation:

Peak value of magnetic field, Bo = 17.9 nT = 17.9 x 10^-9 T

The electromagnetic wave is produced when an oscillating electric and magnetic field interacts each other perpendicularly.

The direction of propagation of electromagnetic wave is perpendicular to both electric and magnetic field.

the relation between the electric field and magnetic field amplitudes is given by

c = \frac{E_{0}}{B_{0}}

where, c be the velocity of light, Eo be the peak value of electric field strength, Bo is the peak value of magnetic field strength.

3\times 10^{8} = \frac{E_{0}}{17.9\times 10^{-9}}}

Eo = 5.37 N/C

4 0
3 years ago
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