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Art [367]
3 years ago
7

A 59.31 kg rock is sitting at the top of a cliff that is 300 m tall. What is the gravitational potential energy of that rock?

Physics
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Answer:

The gravitational potential energy of that rock is 174371.4 J.

Explanation:

Given

  • Mass m = 59.31 kg
  • Height h = 300 m

To determine

We need to find the gravitational potential energy of the rock

We know that the potential energy of a body is termed as the stored energy due to its position.

One kind of energy comes from Earth's gravity — Gravitational potential energy (GPE).

Gravitational potential energy (GPE) can be determined using the formula

GPE = mgh

where

  • m is the mass
  • g is the gravitational acceleration which is equal to g = 9.8 m/s²
  • h is the height
  • GPE is the Gravitational potential energy

now substituting m = 59.31, h = 300 and g = 9.8

GPE = mgh

         =59.31\times 9.8\times 300

         =174371.4 J

Therefore, the gravitational potential energy of that rock is 174371.4 J.

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

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According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

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