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NNADVOKAT [17]
4 years ago
14

Is light one dimensional?

Physics
2 answers:
ruslelena [56]4 years ago
6 0
No, it is not one dimensional since it can act both as a particle and a wave (photon)
kramer4 years ago
3 0

Answer: No.

Explanation: Light exists in 3+1 dimensional space (3 space, 1 time).

You might be interested in
Which of the following best represents a chemical reaction?
Sonja [21]

Answer:

The answer to your question should be D.

Explanation:

reactants are on the laft side of arrow and products are on right side of arrow

7 0
3 years ago
How much water is stored in polar ice caps glaciers and permanent snow
antoniya [11.8K]

Answer:

1.7% is stored in the polar icecaps, glaciers, and permanent snow

Explanation:

70% of the Earth's surface is covered with water; However, only a small percentage, 0.025%, is suitable for human consumption. A limited resource whose demand, according to the forecasts of the Organization for Economic Cooperation and Development (OECD), will skyrocket up to 55% globally between 2000 and 2050.

In the attached infographic we observe that the total volume of water on Earth is 1,386 million cubic kilometers, of which<em> </em><u>1.7% is forming polar ice, cap glaciers and permanent snow</u>.  1,338 million cubic kilometers of that water is in seas and oceans, that is, 96.5% is in seas and oceans, that is, it is salt water that cannot be drunk.

Only 3.5% of the Earth's water is fresh, but of this percentage, only 1% of the Earth's fresh water flows through the river basins in the form of streams and rivers.

8 0
3 years ago
A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o
Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m)   ⇒   h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv²   ⇒   d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0
3 years ago
In aircraft design, the pressure coefficient Cp is usually measured during wind tunnel testing of an aircraft component to predi
Elza [17]

Answer:

Check the explanation

Explanation:

From given data, it can be noted that 95% of given confidently data, means 5% of data is uncertain. According to the question, we have to calculate uncertainty in Cp .

 Kindly check the attached image below for the step by step explanation to the question above.

3 0
3 years ago
Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1
lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

c)\lambda=9.97\times10^{-11}m

d)\lambda=4.03\times10^{-36}m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

\lambda=6.63\times10^{-39}m

c)\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}

\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

λ=∞

6 0
3 years ago
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