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3 years ago

How does the intensity of a sound wave change if the distance from the

Physics

1 answer:

goblinko [34]3 years ago

3 0

Answer:

the answer is that it increases by a factor of 16

Explanation:

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Estimate the total number of bacteria and other prokaryotes in the biosphere of the earth. (assume the bacteria are found to a d

Fofino [41]

Since the Earth is almost spherical in shape, we are actually to find first the volume of the spherical segment at a depth of 1,000 m. The radius of the Earth is 6,371,000 meters. The volume of a spherical segment is:

V = 1/3*πh²(3r - h)
Substituting the values and making sure the units is in mm,
V = 1/3*π(1000 m * 1000 mm/1 m)²[3(6,371,000 m * 1000 mm/1 m) - (1000 m * 1000 mm/1 m)]
V = 2×10²² mm³

Thus, the total amount of bacteria is:

2×10²² mm³ * 100 bacteria/1 mm³ = 2×10²⁴ bacteria

7 0

3 years ago

Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$

specific heat is given as $=\frac{\gamma R}{\gamma -1}$

gas constant =287 J⋅kg−1⋅K−1

$Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}$

specific heat at constant volume

$Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}$

change in internal energy $= Cv(T_2 -T_1)$

$\Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}$

change in enthalapy $\Delta H = Cp(T_2 -T_1)$

$\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}$

change in entropy

$\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})$

$\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})$

$\Delta S = 382.79 J kg^{-1} K^{-1}$

7 0

3 years ago

The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have p

Blizzard [7]

Answer:

True

Explanation:

In this particular case, the area of the graph represents the impulse.

In fact, impulse is defined as the change in momentum of an object:

$I=\Delta p$

Moreover, impulse is also defined as the product between the magnitude of the force acting on an object and the duration of the collision:

$I=F\Delta t$

If we plot a graph of the force versus the time, if the force is constant then this graph will have a rectangular shape, and the area under the graph will simply be the product

$F\cdot \Delta t$

which corresponds to the definition of impulse.

8 0

3 years ago

Dimensionally, which of the following could be a velocity? *

astra-53 [7]

Answer:

5 meters per second

Explanation:

5m is the distance

5m west is the vector

5m per second is the velocity

5m per second west is unknown

5 0

3 years ago

What is the main reason for using a data table for collecting data?

Luba_88 [7]

Answer:

to make calculation more easy to get

Explanation:

if you are using chart or calculate Thermodynamic problems you will not never solve this problem with out using data table for thermodynamic

7 0

3 years ago