Physics, 13.10.2019 11:30 ehklu1226

Ascared elephant has a mass of 7000 kg. the mouse that frightened the elephant is 0.02 kg. the distance between the elephant and the mouse is 1 meter.

a. how much gravitational force does the elephant exert on the mouse?

b. how much gravitational force does the mouse exert on the elephant?

remember to identify all data (givens and unknowns), list equations used, show all your work, and include units and the proper number of significant digits to receive full credit.

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Answer from: nmm32

a) $\rm 9.34 \times 10^-^9\; N$ of gravitational force exerted by the elephent on the mouse.

b) Gravitational force exerted by the mouse on the elephent is,

$\rm F = - 9.34\times 10^-^9\; N$

Given :

M = 7000 kg is the mass of the elephant

m = 0.02 kg is the mass of the mouse

r = 1 m is the separation

Solution:

a) Gravitational force between two objects is given by -

$\rm F = \dfrac{GMm}{r^2}$ ----- (1)

where,

G is the gravitational constant whose value is $6.67\times 10^-^1^1$

M, m are the masses of the two objects

r is the separation between the objects

Now put thw values of G, M, n and r in equation (1) we get,

$\rm F = \dfrac{6.67\times 10^-^1^1\times 7000 \times 0.2}{1^2}$

$\rm F = 9.34 \times 10 ^-^9\;N$

$\rm 9.34 \times 10^-^9\; N$ of gravitational force exerted by the elephent on the mouse.

b) According to Newton's third law of motion -- "When an object A exerts a force (action) on object B, then object B exerts and equal and opposite force (reaction) on object A".

Therefore, gravitational force exerted by the mouse on the elephent is,

$\rm F = - 9.34\times 10^-^9\; N$

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Answer from: xojade

a. $9.34\cdot 10^{-9} N$

The gravitational force between two objects is given by:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

M,m are the masses of the two objects

r is the separation between the objects

In this problem, we have:

M = 7000 kg is the mass of the elephant

m = 0.02 kg is the mass of the mouse

r = 1 m is the separation

Substituting into the equation, we find the gravitational force exerted by the elephant on the mouse:

$F=(6.67\cdot 10^{-11}) \frac{(7000)(0.02)}{1^2}=9.34\cdot 10^{-9} N$

b. $-9.34\cdot 10^{-9}N$

We can solve this part by keeping in mind Newton's third law of motion, which states that:

"When an object A exerts a force (action) on object B, then object B exerts and equal and opposite force (reaction) on object A".

In our problem, we can identify the elephant as object A and the mouse as object B: this means that the gravitational force exerted by the elephant on the mouse is the action, while the gravitational force exerted by the mouse on the elephant is the reaction. The two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the mouse on the elephant is equal to

$F=-9.34\cdot 10^{-9}N$

where the negative sign simply means that the direction is opposite.

Answer from: Quest

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Answer from: Quest

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.

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