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densk [106]
2 years ago
13

A harvest mouse can detect sounds below the threshold of human hearing, as quiet as −10 dB. Suppose you are sitting in a field o

n a very quiet day while a harvest mouse sits nearby. A very gentle breeze causes a leaf 1.5 m from your head to rustle, generating a faint sound right at the limit of your ability to hear it. The sound of the rustling leaf is also right at the threshold of hearing of the harvest mouse. How far is the harvest mouse from the leaf?
Physics
1 answer:
PtichkaEL [24]2 years ago
6 0

Answer:

The distance from harvest mouse to the leaf is 4.74 m.

Explanation:

Given that,

Intensity = -10 dB

Distance = 1.5 m

We need to calculate the power of intensity

Using formula of power of intensity

P=I_{0}A

P=I_{0}\times4\pi r^2

Put the value into the formula

P=1.0\times10^{-12}\times4\pi\times(1.5)^2

P=28.27\times10^{-12}\ W/m^2

We need to calculate the minimum intensity level

Using formula of minimum intensity

I=10\log_{10}(\dfrac{I}{I_{0}})

-10=10\log_{10}(\dfrac{I}{I_{0}})

\log_{10}\dfrac{I}{I_{0}}=-1

I=10^{-1}\times(1.0\times10^{-12})

I=1\times10^{-13}\ W/m

We need to calculate the area

Using formula of intensity

I=\dfrac{P}{A}

A=\dfrac{P}{I}

Put the value into the formula

A=\dfrac{28.27\times10^{-12}}{1\times10^{-13}}

A=282.7\ m^2

We need to calculate the distance

Using formula of area

A=4\pi r^2

Put the value into the formula

282.7=4\pi\times r^2

r^2=\dfrac{282.7}{4\pi}

r=\sqrt{\dfrac{282.7}{4\pi}}

r=4.74\ m

Hence, The distance from harvest mouse to the leaf is 4.74 m.

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A supertanker filled with oil has a total mass of 6.1 x108 kg. If the dimensions of the ship are those of a rectangular box 300
IrinaVladis [17]

Answer:

The bottom of the sea is 25 m below sea level.

Explanation:

Given data

Mass = 6.1 × 10^{8} \ kg

\rho_{sea} = 1020\  \frac{kg}{m^{3} }

We know that Buoyant force on the tank is equal to gravity force of the tank.

F_B = F_g

(\rho_{Fluid}) (g) (V_{disp}) = m g

(\rho_{Fluid})  (V_{disp}) = m

1020 × V_{disp} = 6.1 × 10^{8}

V_{disp} = 598039.21 m^{3}

We know that

V_{disp} = W × L × H

598039.21 = 300 × 80 × H

H = 25 m

Therefore the bottom of the sea is 25 m below sea level.

7 0
3 years ago
Two capacitors have the same size of plates and the same distance 4 mm between the plates The potentials of the two plates in ca
OverLord2011 [107]

Answer:

    E = -4000 N / C

Explanation:

The potential and electric field are related

         V = - E s

          E = - V / s

we reduce the magnitudes to the SI system

          s = 4 mm (1 m / 1000 mm) = 0.004 m

we calculate

          E = - 16 /0.004

          E = -4000 N / C

8 0
3 years ago
Which of the following best describes the suns size compared with other stars?
Rufina [12.5K]
1) D: enormous
2) D: gravity
8 0
3 years ago
Consider the system consisting of two blocks connected by a rope and a pulley. The coefficient of static friction between the ra
RSB [31]

Answer:

1.2 kg

Explanation:

Let UP ramp be the positive direction

                                                  F = ma

   T     -   Wt ||      -         Ff            = m(0)

  mg   -  Μgsinθ -      μΜgcosθ    = 0

m(9.8) - 13sin35 - 0.36(13)cos35 = 0        

                                                 m = 13(sin35 + 0.36cos35) / 9.8

                                                 m = 1.15205... ≈ 1.2 kg

5 0
2 years ago
A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one
alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R=13\Omega

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper  wire=Sum of resistance of two piece of wires

r+7r=13

8r=13

r=\frac{13}{8}ohm

Resistance of long piece of wire=7\times\frac{13}{8}=\frac{91}{8}\Omega

Resistance of short piece of wire =\frac{13}{8}\Omega

Resistivity of wire and cross section area of wire remains same .

Let L be the total  length  of wire and L' be the length of short  piece of wire.

We know that

R=\frac{\rho L}{A}=\frac{\rho}{A}L=KL

\frac{R}{L}=K

Where K=\frac{\rho}{A}=Constant

Using the formula

\frac{13}{L}=\frac{\frac{13}{8}}{L'}

\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}

L'=\frac{L}{8}

Length of short piece of wire=L'=\frac{L}{8}

Length of long piece of  wire=L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L

% of length of short piece of   wire=\frac{\frac{L}{8}}{L}\times 100=12.5%%

The resistance of the short piece=\frac{13}{8}\Omega

The resistance of the long piece=\frac{91}{8}\Omega

8 0
3 years ago
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