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yanalaym [24]
3 years ago
15

Why don’t we feel the gravitational force of a large object such as a skyscraper semi-truck?

Physics
1 answer:
Kobotan [32]3 years ago
8 0

Answer:

Se the explanation below

Explanation:

We do not feel these forces of these bodies, because they are very small compared to the force of Earth's attraction. Although its mass is greater than that of a human being, its mass is not compared to the Earth's mass. In order to understand this problem we will use numerical data and the universal gravitation formula, to give validity to the explanation.

<u>Force exerted by the Earth on a human being</u>

<u />

F=G*\frac{m_{1}*m_{2}}{r^2}

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 5.97*10^24[kg]

r = distance from the center of the earth to the surface or earth radius = 6371 *10^3 [m]

<u />

Now replacing we have

F = 6.673*10^{-11} *\frac{80*5.97*10^{24}}{(6371*10^{3})^{2}  } \\F = 785[N]

<u>Force exerted by a building on a human being</u>

<u />

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 300000 [ton] = 300 *10^6[kg]

r = distance from the building to the person = 2[m]

F = 6.673*10^{-11}*\frac{80*300*10^6}{2^{2} }  \\F= 0.4 [N]

As we can see the force exerted by the Earth is 2000 times greater than that exerted by a building with the proposed data.

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80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting te
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the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

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mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

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let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C

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