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4 years ago

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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Anvisha [2.4K]

Answer:

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Explanation:

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3 years ago

the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance

Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

$F = G\frac{Mm}{r^{2} }$

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = $\frac{1}{4}$ G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

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F₂ = $\frac{1}{4}$ F₁

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3 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

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So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

3 years ago