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Wittaler [7]
3 years ago
9

a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on

the earth surface.
Physics
1 answer:
alekssr [168]3 years ago
4 0

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}

Now, if the body is on the surface of the Earth, its weight (w) would be:

F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\

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Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

Given:

mass of first particle, m_1=6.7\ kg

mass of second particle, m_2=5.1\ kg

mass of third particle, m_3=3.7\ kg

coordinate position of first particle in meters, (x_1,y_1)\equiv(4.2,0)

coordinate position of second particle in meters, (x_2,y_2)\equiv(0,2.8)

coordinate position of third particle in meters, (x_3,y_3)\equiv(0,0)

<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

F_{32}=16.05\times 10^{-11}\ N

towards positive X axis.

<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

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Answer:

–77867 m/s/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

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Time (t) = 0.00075 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

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Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

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