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Sati [7]
3 years ago
15

A sound wave is emitted into the water from the bottom of a boat. What will happen to the sound wave as it travels through the w

ater?
A. It will slowly lose energy to the water around and become weaker the farther it goes

B. It will absorb thermal energy from the water and speed up until it reaches a maximum speed

C. It will continue along its original path at the same speed idefinitly.

D. It will not travel more than 10 m due to the density of water
Biology
2 answers:
viva [34]3 years ago
4 0
It is is A but i could be wrong
scoray [572]3 years ago
3 0

Answer:

A. It will slowly lose energy to the water around and become weaker the farther it goes

Explanation:

Speed of sound in a medium

c = √(K/ρ)

K is the elastic bulk modulus and ρ is the density

For water

K = 2.2 GPa

ρ = 997 kg/m³

c = √(2.2×10⁹/997)

c = 1485.46 m/s

For material of boat (say steel)

K = 167 GPa

ρ = 8050 kg/m³

c = √(167×10⁹/8050)

c = 4554.71 m/s

Here it can be seen the speed in solid is faster than in water so sound will lose energy when it passes through water.

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This is an interesting problem involving astronomy, in fact, simple physics.
Let r=distance of sun to mars, in metres

<span>Mars had an orbital period of 1.88 years.
=>
tangential velocity, v, of the planet, in m/s is
v=\frac{2\pi{r}}{T}
</span>=\frac{2\pi{r}}{1.88*365.25*86400} m/s, accounting for leap years
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        where m=mass of mars = 6.39*10^(24) kg
=\frac{mv^2}{r}
=\frac{6.39*10^{24}v^2}{r}
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The gravitation pull from the sun, Fg, is given by
Fg=\frac{GMm}{r^2}
    where G=grav. const., =6.67408*10^(-11) m^3<span> kg^(</span>-1)<span> s^(</span>-2)
              M=mass of sun=1.989*10^(30) kg
=\frac{6.67408*10^{-11}1.989*10^{30}6.39*10^{24}}{r^2}
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Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
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7.26168*10^9\pi^2r=\frac{8.4826*10^44}{r^2}
Solving for the real root:
r^3=\frac{8.48256*10^44}{7.26168*10^9*%pi^2}
=\frac{1.1681263*10^{35}}{\pi^2}
=2.279*10^11 m


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