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Ivenika [448]
3 years ago
14

1. Find the magnitude of the gravitational force a 66.5 kg person would experience

Physics
1 answer:
nalin [4]3 years ago
8 0

Answer:

Gravitational forces: Earth: W = 650.969\,N, Mars: W = 246.449\,N, Pluto: W = 38.504\,N

Explanation:

The weight (W), measured in newtons, experimented by the person on the surface of the planet is defined by Newton's Laws of Motion:

W = m\cdot g (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

From Newton's Law of Gravitation we derive this expression for gravitational acceleration:

g = \frac{G\cdot M}{R^{2}} (2)

Where:

G - Gravitational constant, measured in cubic meters per kilogram-square second.

M - Mass of the planet, measured in kilograms.

R - Radius of the planet, measured in meters.

By (2) we calculate the gravitational acceleration for each planet:

Earth (G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.97\times 10^{24}\,kg, R = 6.38\times 10^{6}\,m)

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.97\times 10^{24}\,kg)}{(6.38\times 10^{6}\,m)^{2}}

g = 9.789\,\frac{m}{s^{2}}

Mars (G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 6.42\times 10^{23}\,kg, R = 3.40\times 10^{6}\,m)

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (6.42\times 10^{23}\,kg)}{(3.40\times 10^{6}\,m)^{2}}

g = 3.706\,\frac{m}{s^{2}}

Pluto (G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 1.25\times 10^{22}\,kg, R = 1.20\times 10^{6}\,m)

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1.25\times 10^{22}\,kg)}{(1.20\times 10^{6}\,m)^{2}}

g = 0.579\,\frac{m}{s^{2}}

Lastly, we calculate the gravitational force for each case by (1):

Earth (m = 66.5\,kg, g = 9.789\,\frac{m}{s^{2}})

W =(66.5\,kg)\cdot \left(9.789\,\frac{m}{s^{2}} \right)

W = 650.969\,N

Mars (m = 66.5\,kg, g = 3.706\,\frac{m}{s^{2}})

W = (66.5\,kg)\cdot \left(3.706\,\frac{m}{s^{2}} \right)

W = 246.449\,N

Pluto (m = 66.5\,kg, g = 0.579\,\frac{m}{s^{2}})

W = (66.5\,kg)\cdot \left(0.579\,\frac{m}{s^{2}} \right)

W = 38.504\,N

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1) v₀x = 13.76 m/s

2) v₀y = 19.66 m/s

3) ymax = 21.199 m

4) X = 55.1746 m

5) and 6) y = 18.4 m

Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m

4) We can use this equation

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where y = y₀ = 1.5 m

then

1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²

⇒   0.02588 x² - 1.42815 x = 0

Solving this equation we get

x₁ = 0     and    x₂ = 55.1746 m

The distance between the two girls is 55.1746 m

5) and 6) If   v₀x = 15 m/s = vx   and   ymax = 24 m

y = ?   when x = (xmax/2)

ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))

⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s

then we get α' as follows

α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s

Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

⇒  y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))

⇒  y = 18.4 m

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