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zmey [24]
1 year ago
13

If the distance to a point source of sound is doubled, by what multiplicative factor does the intensity change?

Physics
1 answer:
alina1380 [7]1 year ago
8 0

If the distance to a point source of sound is doubled, by a multiplicative factor of 4, the intensity changes.

Intensity of sound is the sound which is perpendicular to sound wave propogation per unit area. It is dependent on the Surface of source sound.

Intensity is the Power per unit area. Its SI unit is Watt/m².

As we move away from a source of sound, the sound starts to diminish. This is due to the decreasing sound intensity with distance.

It can also be understood by the fact that on increasing distance, the Power radiated by the source spreads over a larger area. Hence, the Intensity decreases gradually.

Since, Intensity is proportional to the square of the distance.

Hence, on doubling the distance, Intensity reduces to one fourth of the initial intensity or reduces by a multiplicative factor of 4.

Learn more about Intensity here, brainly.com/question/17583145

#SPJ4

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What is the unit of k (spring constant) in SI system?
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Answer:

SI unit of k (spring constant) = N/m

Explanation:

We have expression for force in a spring extended by x m given by

         F = kx

Where k is the spring constant value.

Taking units on both sides

         Unit of F = Unit of k x Unit of x

          N = Unit of k x m

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SI unit of k (spring constant) = N/m

7 0
3 years ago
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Which statement best defines an electric field?
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A region around a charged particle or object within which a force would be exerted on other charged particles or objects
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Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
3 years ago
An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel
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Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

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∴                 g = 2S/t²

Substituting the given values,

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Hence, the gravitational acceleration of the planet is, g = 8 m/s²

7 0
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Ksivusya [100]

Answer : Radius of Earth is 6,340 Km.

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We know that the relation between centripetal acceleration and tangential velocity is :

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v is tangential velocity

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r=6340909\ m

or

r=6,340\ Km

The Earth's radius is 6,340 Km. Hence, this is the required solution.

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