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maks197457 [2]
3 years ago
15

Is sound faster than the speed of light ?​

Physics
2 answers:
Elden [556K]3 years ago
7 0

Answer:

no

Explanation:

light is faster than sound. we can go past the speed of sound but not the speed of light.

sound travels as a wave and deminishes over time due to having to travel through all of the atoms of the things it passes.

light behaves as a wave and a particle and refracts off of each thing that it touches (what allows us to see color)

IRISSAK [1]3 years ago
6 0

False..................

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What keeps both the cars pressed down on the road? ​
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Answer:

Gravity

Explanation:

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3 years ago
On sunday a strong gale blows near beth’s house. the best estimate for the wind speed is ____ miles/hour.
Rufina [12.5K]

Answer:

B. 47-54 miles/hour

Explanation:

Gale is a strong wind which is depicted by red warning flag. According to U.S. National Weather Service gale is a sustained surface wind. It is also used to refer winds from tropical coastal areas.

On the basis of force of wind gale is divided in four groups:

  1. Near gale   - 32-38 mph
  2. Gale            - 39-46 mph
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3 years ago
What is the difference between organic and inorganic material?
babymother [125]

Organic materials comes from living things while inorganic materials comes from non living things


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2 years ago
Electromagnetic waves differ fundamentally from either water or sound waves because they
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8 0
3 years ago
A sample of gas with a volume of 750 ml exerts a pressure of 98 kpa at 30◦c. What pressure will the sample exert when it is comp
Tanzania [10]

Answer:

241 kPa

Explanation:

The ideal gas law states that:

pV=nRT

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

We can rewrite the equation as

\frac{pV}{T}=nR

For a fixed amount of gas, n is constant, so we can write

\frac{pV}{T}=const.

Therefore, for a gas which undergoes a transformation we have

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where the labels 1 and 2 refer to the initial and final conditions of the gas.

For the sample of gas in this problem we have

p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K

So we can solve the formula for p_2, the final pressure:

p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa

4 0
3 years ago
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