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Lerok [7]
4 years ago
9

What is the speed of sound in air when the temperature is 20°C?

Physics
2 answers:
Anarel [89]4 years ago
5 0
The answer to your question is 343 m/s
Minchanka [31]4 years ago
4 0

Answer:

The speed of sound in air when the temperature is 20°C is 343 m/s.

Explanation:

Given that,

Temperature T = 20°C = 20+273=293 K

The speed of sound in air is the distance traveled divided by time.

We know that,

The relation between the speed of sound in air and temperature is defined as:

v= 331\sqrt{\dfrac{T}{273}}

Where, T = temperature in Kelvin

v = speed of sound in air

We substitute the value into the formula

v = 331\sqrt{\dfrac{293}{273}}

v = 342.9 = 343\ m/s

The speed of sound in air is 331 m/s at 0° C, 343 m/s at 20°C and 346 m/s at 25°C.

Hence, the speed of sound in air when the temperature is 20°C is 343 m/s.

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A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

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Answer:

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Round to three significant digits

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