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ASHA 777 [7]
3 years ago
7

The gravitational self potential energy of a solid ball of mass density ρ and radius R is E. What is the gravitational self pote

ntial energy of a ball of mass density ρ and radius 2R in terms of E?
Physics
1 answer:
Alenkasestr [34]3 years ago
8 0
It will be
E = mgh.
where h and g are constant thus
m can be written as 4/3πr^3*density
E = 4/3πr^3* density
E? = 4/3π(2R)^3* density
= 4/3π8r^3
thus the e will be 4/3π8r^3* density/4/3πr^3*density nd thus you get 8E ..
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Furkat [3]

The correct answer is:

Product of their masses


In fact, the gravitational pull between two objects is given by:

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From the equation, we immediately see that the gravitational attraction is directly proportional to the product of the masses.

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3–101 It is estimated that 90 percent of an iceberg’s volume is below the surface, while only 10 percent is visible above the su
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The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³

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Given that,

Volume of seawater = 90 % of Volume of iceberg

Volume of visible surface = 10 %

Density of seawater = 1025 kg/m³

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Using equilibrium condition

W_{I}=F_{B}

\rho_{I}\times V_{I}\times g= \rho_{w}\timesV_{w}\times g

Put the value into the formula

\rho_{I}\times V_{I}\times g=\rho_{w}\times0.9V_{I}\times g

\rho_{I}=1025\times0.9

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We need to calculate the density of the iceberg

Using equilibrium condition

W_{I}=F_{B}

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Put the value into the formula

\rho_{I}\times V_{I}\times g=\rho_{w}\times0.1V_{I}\times g

\rho_{I}=1025\times0.1

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5 0
3 years ago
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Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.

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sleet_krkn [62]

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B. The answer is given below.

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