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ella [17]
3 years ago
5

How does the latent heat of fusion of water help slow the decrease in air temperature,perhaps preventing temperatures from falli

ng significantly below 0◦C, in the vicinity oflarge bodies of water?
Physics
1 answer:
love history [14]3 years ago
6 0

Answer:

Water has the ability to release a large amount of energy during the freezing process. All of this energy is released into the air, resulting in a greater movement of air particles that will increase the air temperature. This heat will prevent air temperatures from drastically falling below 0ºC.

Explanation:

The latent heat of water occurs when the water is changing its physical state. In other words, when substances are changing their physical state, the amount of heat calculated for this change is called latent heat, and as we have already said, this heat is not related to the heat exchanges between two systems, but the change of state. physicist.

When water is changing from its liquid state to a solid state, we call latent heat latent heat from fusion (this is because fusion is the name we give to when liquid water is turning to ice). In the process of freezing the page, latent heat releases a large amount of energy into the air near the water. This energy agitates the air molecules, generating heat and preventing the air from reaching temperatures below 0ºC.

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Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

7 0
2 years ago
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the pow
allsm [11]

This question is incomplete, the complete question;

you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity

Options;

a) 200 nm; 0.9 mW

b) 100 nm, 0.0059 mW

c) 200 nm; 0 mW

d) 100 nm; 0.9 mW

e) 200 nm; 0.0059 mW

Answer:

the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

Explanation:

Given that the instrument here is an interferometer.

Maximum intensity is obtained when the two waves are exactly in phase.

that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.

The phase factor of this point is taken as ∅ = 0

Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π

This corresponds to a path difference between the two waves as half of the wavelength. λ/2

The light gets reflected from the mirror.

Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)

hence, to get a path difference of λ/2 the mirror should move half of this distance only

so, the mirror should move;

l = λ/4

here, wavelength is 400nm

the length moved by the mirror = 400/4 = 100 nm

The intensity is given by the equation;

l = l1 + l2 + 2√l1l2cos(∅)

where

l1 = 2.25 mW

l2 = 2.025 mW

∅ = π

so we substitute

l = 2.25 + 2.025 - 2√(2.25 × 2.025)

l = 4.275 - 4.26907

l = 0.0059

Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer  

5 0
2 years ago
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