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alina1380 [7]
3 years ago
8

What is the speed of sound in air with a temperature of 25 degrees C​

Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0

Answer:

The speed of sound through air is approximately 343 m/s at normal room temperature, which is at 20 °C. The speed of sound through air is 346 m/s at 25 °C.

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Due Ma<br> duart<br> ded<br> out<br> 25 N<br> 35 N<br> 1-03
Anuta_ua [19.1K]

Answer:

what's that all about

hehehwhe

Explanation:

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8 0
2 years ago
Kyle has a mass of 54kg and is moving at 3 m/s what is his kinetic energy
Mars2501 [29]

Answer:

243J

Explanation:

K.E = 1/2 x 54 x 3^2

K.E = 1/2 x 54 x 9

K.E = 1/2 x 486

K.E = 486/2

K.E = 243J

6 0
3 years ago
A car is traveling north at 17.7 m/s . After 6 it’s velocity is 141 in the same direction. Find the magnitude and direction of t
Furkat [3]

By equation of motion we have   v = u + at

Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration

Here v = 141 m/s, u = 17.7 m/s and t = 6 s

On substitution we will get

        141 = 17.7+ 6a

       So, a = (141-17.7)/6 = 20. 55 m/s^{2}

       Aceeleration = 20. 55 m/s^{2} along north direction.


3 0
3 years ago
As longitudional waves travel, particles in the medium are pushed together and then pulled apart. We call this
andreyandreev [35.5K]
Compression and rarefraction, the other guy's answer it's wrong
4 0
2 years ago
Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 time
skad [1K]

Answer:

g' = 13.5 m/s²

Explanation:

The acceleration due to gravity on surface of earth is given by the formula:

g = GMe/Re²   --------------- euation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

Me = Mass of Earth

Re = Radius of Earth

Now, the the acceleration due to gravity on the surface of Kepler-62e is:

g' = GM'/R'²   --------------- euation 1

where,

g' = acceleration due to gravity on surface of Kepler-62e

G = Universal Gravitational Constant

M' = Mass of Kepler-62e = 3.57 Me

R' = Radius of Kepler-62e = 1.61 Re

Therefore,

g' = G(3.57 Me)/(1.61 Re)²

g' = 1.38 GMe/Re²

using equation 1:

g' = 1.38 g

where,

g = 9.8 m/s²

Therefore,

g' = 1.38(9.8 m/s²)

<u>g' = 13.5 m/s²</u>

6 0
3 years ago
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