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Zanzabum
3 years ago
12

Does laser refract through water or reflect ?

Physics
2 answers:
Novosadov [1.4K]3 years ago
7 0

Answer:

Calm water looks like a mirror from below because underwater light is entirely reflected when it comes from beyond a certain angle. This so-called total internal reflection also allows low power lasers to bend the waters surface.

thank you

pls mark as BRAINLIST

NemiM [27]3 years ago
3 0

Answer:

If you shine a light straight down at water, most of the light will go into the water, and some will reflect.

SO that would means:

- Laser would refract throught the water (in some part of it)

- But laser would also reflect from the water surface (in another part of it )

Another easy example:

- Sunshine go through water? (yes in the first few meters)

- Sunshine reflect with water? (yes, most of them would reflect )

Hope this helped :3

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A demented scientist creates a new temperature scale, the "Z scale." He decides to call the boiling point of nitrogen 0°Z and th
slavikrds [6]

Answer:

A) Z = 0.577 C +112.931

Z = 0.577*(100) +112.931=170.631 Z

B) C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C

C) K = C +273.15

K = -22.41 +273.15 =250.739 K

Explanation:

For this case we want to create a function like this:

Z = a C + b

Where Z represent the degrees for the Z scale C the Celsius grades and  tha valus a and b parameters for the model.

The boiling point of nitrogen is -195,8 °C

The melting point of iron is 1538 °C

We know the following equivalences:

-195.8 °C = 0 °Z

1538 °C = 1000 °Z

Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)

So then we can calculate the slope for the linear model like this:

a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}

And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

0 = 0.577 (-195.8) + b

And if we solve for b we got:

b = 0.577*195.8 =112.931 Z

So then our lineal model would be:

Z = 0.577 C +112.931

Part A

The boiling point of water is 100C so we just need to replace in the model and see what we got:

Z = 0.577*(100) +112.931=170.631 Z

Part B

For this case we have Z =100 and we want to solve for C, so we can do this:

Z-112.931 = 0.577 C

C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C

Part C

For this case we know that K = C +273.15

And we can use the result from part B to solve for K like this:

K = -22.41 +273.15 =250.739 K

6 0
3 years ago
A girl and boy pull in opposite directions on a stuffed animal. The girl exerts a force of 3.5 N. The mass of the stuffed animal
givi [52]
I'm gonna have to assume the girl is on the right side and boy on left.
The net force is the sum of all forces on an object (includes negatives).
Let's say the force of the boy is variable <em>b</em>.  Use the formula F = ma.

<em>b </em>+ 3.5 = 0.2(2.5)

This is now simple algebra.  Solve to get that <em />the boty is exerting a force of -3N to the left.
5 0
3 years ago
Read 2 more answers
3 PHYSICAL SCIENCE: In order to figure out what is in a sealed box, you perform some tests. The box seems heavy for its size and
rjkz [21]
Some sort of magnetic metal

Metals are heavier per cubic unit than other materials such as air or water, and also are much more magnetic than other materials
7 0
4 years ago
What is the SI (metric) unit of FORCE?<br><br> A. meter<br> B. newton
Tresset [83]

What is the SI (metric) unit of FORCE?

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with symbol ( N )

All the best !

4 0
3 years ago
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A ball of mass 5 kg attached to a string is swung in a horizontal circle of radius 0.5 m. If the tension in the string is 10 N,
kirill115 [55]

Answer:

0 J

Explanation:

given,

mass of the ball = 5 kg

radius of the horizontal circle = 0.5 m

tension in the string = 10 N

Work done = ?

Work done by the tension in the circular path will be equal to zero.

This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.

so, work done = F s cos θ

     θ = 90°,

work done = F s cos 90°        ∵ cos 90° = 0

Work done = 0 J

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