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schepotkina [342]

3 years ago

7

A echo occurs when sound does what?

2 answers:

klio [65]3 years ago

8 0

Echoes occur when a reflected soundwave reaches the ear more than 0.1 seconds after the original sound wave was heard. ... There will be an echoinstead of a reverberation. Reflection of sound waves off of surfaces is also affected by the shape of the surface.

(mark as brainly please)

kirza4 [7]3 years ago

7 0

An echo occurs when a sound wave bounces off of a smooth surface and travels back in the same direction it came (reflects).

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A pail of water is rotated in a vertical circle of radius r = 1.6 m . The acceleration of gravity is 9.8 m/s 2 . r v What is the

Snowcat [4.5K]

Answer:3.95 m/s

Explanation:

Given

radius of circle $r=1.6 m$

Water will not spill if the centripetal acceleration is greater than acceleration due to gravity

i.e. $a_c\geq g$

$a_c=\frac{v^2}{r}$

$\frac{v^2}{r}\geq g$

$v_{min}=\sqrt{gr}$

$v_{min}=\sqrt{9.8\times 1.6}$

$v_{min}=\sqrt{15.68}$

$v_{min}=3.95 m/s$

8 0

3 years ago

A 5 m3 tank containing 5kg of an unknown ideal gas at 500 kPa is connected through a valve to another tank containing 10 kg of t

Ivan

Answer:

a) $V_{T} = 9\,m^{2}$ , b) $m_{T} = 15\,kg$ , c) $P_{T} = 416.667\,kPa$

Explanation:

a) The equation of state for ideal gas is:

$P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

Given the existence of an isothermal process, the following relation is derived:

$\frac{P_{1}\cdot V_{1}}{m_{1}} = \frac{P_{2}\cdot V_{2}}{m_{2}}$

The volume of the other tank is:

$V_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}$

$V_{2} = \left(\frac{10\,kg}{5\,kg} \right)\cdot \left(\frac{200\,kPa}{500\,kPa}\right)\cdot (5\,m^{3})$

$V_{2} = 4\,m^{3}$

The total volume is:

$V_{T} = V_{1} + V_{2}$

$V_{T} = 5\,m^{3} + 4\,m^{3}$

$V_{T} = 9\,m^{2}$

b) The total mass is:

$m_{T} = m_{1} + m_{2}$

$m_{T} = 5\,kg + 10\,kg$

$m_{T} = 15\,kg$

c) The pressure of the gas in the two tanks is:

$P_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{V_{1}}{V_{2}}\right)\cdot P_{1}$

$P_{T} = \left(\frac{15\,kg}{5\,kg}\right)\cdot \left(\frac{5\,m^{2}}{9\,m^{2}} \right)\cdot (500\,kPa)$

$P_{T} = 416.667\,kPa$

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