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charle [14.2K]
3 years ago
11

The inverse square law applies to:

Physics
2 answers:
Serga [27]3 years ago
8 0

Answer:

Intensity of sound

Gravitational Force

Intensity of light

Electric Force

Explanation:

As we know that inverse square law will depends on the square of the distance inversely

So for

Intensity of sound or intensity of Light for very small size source we can say intensity will be defined as power per unit area.

So we can say

I = \frac{P}{4\pi r^2}

Now similarly we know that gravitational force is given as

F = \frac{Gm_1m_2}{r^2}

also we know that electrostatic force is given as

F = \frac{kq_1q_2}{r^2}

so all above are following inverse square law

Roman55 [17]3 years ago
4 0

There are more correct choices on the list than incorrect ones.

The inverse square law applies to

-- the intensity of sound from a small source

-- the Gravitational Force

-- the Intensity of light from a small source

-- the Electric Force  from a point charge or a small charge.

It doesn't apply to the Magnetic Force.

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Answer:

T = 1010 degree Celsius

Explanation:

mass of ball (Mb) = 100 g

mass of water (Mw) = 400 g

temp of water = 0 degree

specific heat of platinum (C) = 0.04 cal/g degree celsius

we can calculate the temperature of the furnace from the equation before

Mb x C x (temp of furnace (T) - equilibrium temp) = Mw x (equilibrium temp - temp of furnace)

100 x 0.04 x ( T - 10) = 400 x (10 - 0)

4 (T - 10) = 4000

T - 10 = 1000

T = 1010 degree Celsius

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Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object
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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
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Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

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3 years ago
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