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Tanzania [10]
3 years ago
15

Explain the conditions in which a person can or can’t hear a echo

Physics
2 answers:
victus00 [196]3 years ago
6 0

Answer:

Explained

Explanation:

The necessary condition for an echo to occur are

1) The distance between the sound source and the object of reflection must be equal to or greater than 17 meters.

2) the time period between hearing the original sound and the echo must not be less than 0.1 second, because human ear is not that sensitive to distinguish between two sounds within a gap of time less than 0.1 second.

Olenka [21]3 years ago
5 0

When your in the mountains or in a quiet room

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rosijanka [135]

A and B are equivalent.  That's one way instruments are often grouped. (the "sopranos", the "altos", the "bass")

C is another way instruments are often grouped; (the "woods", the "brass")

D is another way instruments are often grouped; (the "strings", the "percussions")

6 0
3 years ago
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The height of the Washington Monument is measured to be 170 m on a day when its temperature is 35.0°C. What will the change in i
Alecsey [184]

Answer:

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m

Explanation:

Thermal coefficient of marble varies between (5.5 - 14.1) ×10⁻⁶/K = α

So, let us take the average value

(5.5+14.1)/2 = 9.8×10⁻⁶ /K

Change in temperature = 35-(-18) = 53 K = ΔT

Original length = 170 m = L

Linear thermal expansion

\frac{\Delta L}{L} = \alpha\Delta T\\\Rightarrow \Delta L=\frac{\alpha\Delta T}{L}\\\Rightarrow \Delta L=9.8\times 10^{-6}\times 53\times 170

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m (subtraction because of cooling)

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3 years ago
How does exhailing remove waste from the body? Explain the systems that make this happen‚ using complete sentences​
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Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea
steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at  P = 100 k Pa and T = 500 K

mass is constant

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\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}

Pressure is same because it's not changing due to process

\dfrac{V}{500} = \dfrac{2 V}{T_2}

T_2 = 1000\ K

\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}

\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})

m = \dfrac{P_1V_1}{RT_1}

m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}

m = 1.39 x 10⁻³ Kg

\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)

\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K

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