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Jet001 [13]
3 years ago
12

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

? (Use any variable or symbol stated above as necessary.)'
Physics
1 answer:
Margarita [4]3 years ago
4 0

Answer:

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

Explanation:

  • We have to make a hollow sphere of inner  radius r_1 and outer radius r_2.

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

V_t=\frac{4}{3} \pi.r_2^3

And the volume of the hollow space in the sphere:

V_h=\frac{4}{3} \pi.r_1^3

Therefore the net volume of material required to make the sphere:

V=V_t-V_h

V=\frac{4}{3} \pi(r_2^3-r_1^3)

  • Now let the density of the of the material be \rho.

<u>Then the mass of the material used is:</u>

m=\rho.V

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

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In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 350MV . Th
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Complete Question

In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 350 MV (35,000,000 V). The bottoms of the thunderclouds are typically 1500 m above the earth, and can have an area of 120 km^2. Modeling the earth/cloud system as a huge capacitor, calculate

a. the capacitance of the earth-cloud system

b. the charge stored in the "capacitor"

c. the energy stored in the capacitor

Answer:

a

 C =  7.08 *10^{-7} \  F

b

  Q =  24.78 \  C

c

 E =433650000 \ J

Explanation:

From the question we are told that

The potential difference is  V  =  35000000 V

The distance of the bottom of the thunderstorm from the earth is  d = 1500 m

The area is  A =  120 \  km^2 =  120 *10^{6} \  m^2

Generally the capacitance of the earth cloud system is mathematically represented as

         C =  \epsilon_o *  \frac{A}{d}

 Here \epsilon_o is the permitivity of free space with as value \epsilon_o =  8.85 *10^{-12} \  C/(V\cdot m)

So

     C =  8.85*10^{-12} *  \frac{120*10^{6}}{1500}

=>  C =  7.08 *10^{-7} \  F

Generally the charge stored in the capacitor (earth-cloud system) is mathematically represented as

       Q =  C  *  V

=>    Q =  7.08 *10^{-7}  *   35000000

=>    Q =  24.78 \  C

Generally the energy stored in the capacitor is mathematically represented as

       E = \frac{1}{2}  * Q *  V

=>    E = \frac{1}{2}  *   24.78 *  35000000

=>    E =433650000 \ J

   

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