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agasfer [191]
2 years ago
8

The frequency of a certain sound is 440 Mz. What is the wavelength of this sound when the temperature of the air is (a) 20°C; (b

) 30°C
Physics
1 answer:
Serggg [28]2 years ago
3 0

Answer:

Explanation:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for

λ

:

λ

=

v

f

Let's plug in our given values and see what we get!

λ

=

340

m

s

440

s

−

1

λ

=

0.773

m

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2 years ago
The Grand Canyon is 1800 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of
snow_tiger [21]

Answer:

Explanation:

Given

height of grand canyon is h=1800\ m

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u=0

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h=height

u=initial velocity

a=acceleration

t=time

here a=acceleration due to gravity

1800=0+\frac{1}{2}\times 9.8\times t^2

t^2=\frac{2\times 1800}{9.8}

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3 years ago
A monk is sitting atop a mountain in complete rest in meditation. What is the kinetic Energy of the monk? (assume mass of 65 kg
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Explanation:

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3 years ago
Ballon volume of 3200ml of xenon gas is at a gauge pressure of 122kPa and a temperature of 27c. What is the volume when the ball
N76 [4]

Given:

The initial volume of the gas, V₁=3200 ml=3.2×10⁻³ m³

The initial pressure of the gas, P₁=122 kPa

The initial temperature of the gas, T₁=27 °C=300 K

The final temperature, T₂=65 °C=338 K

The final pressure, P₂=112 kPa

To find:

The final volume of xenon gas.

Explanation:

From the combined gas law,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Where V₂ is the volume after it is heated.

On rearranging the above equation,

V_2=\frac{T_2P_1V_1}{T_1P_2}

On substituting the known values,

\begin{gathered} V_2=\frac{338\times112\times10^3\times3.2\times10^{-3}}{300\times112\times10^3} \\ =3.61\text{ m}^3 \end{gathered}

Final answer:

The volume of the balloon when it is heated is 3.61 m³

4 0
1 year ago
Consider a piston filled with 3 mols of an ideal gas, kept at a constant temperature 290 K. We slowly compress the gas starting
kap26 [50]

Answer: -5013.65\ J

Explanation:

Given

No of moles n=3

Temperature T=290\ K

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Final volume V_2=1\ m^3

Work done in constant temperature process is

W=nRT\ln \left(\dfrac{V_2}{V_1}\right)

Insert the values

\Rightarrow W=3\times 8.314\times 290\ln \left (\dfrac{1}{2}\right)\\\\\Rightarrow W=-870\times 8.314\times \ln (2)\\\Rightarrow W=-5013.65\ J

5 0
3 years ago
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