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mr Goodwill [35]
3 years ago
6

You and you friend are sitting on the floor in an empty room a distance, d, apart, and you both have an equal mass. Assume the o

nly force acting on you and your friend is the gravitational force you each exert on each other. No other forces are present.
a. How long will it take due to gravitational forces before you and your friend meet directly in the middle? Note: You may assume the acceleration remains constant throughout the motion.
b. In reality, the acceleration is not constant and is continually changing. Why is this?
c. If the acceleration is changing, what effect will this have on the derived time? Will the time increase, decrease, or stay the same? Why?
Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:ppppppppooooooooooopppppppppppp

Explanation:

pppppppppooooooooooopppppppppppp

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zepelin [54]

Answer:

V= 2.7 [v]

Explanation:

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3 years ago
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Answer:

Explanation:

Its not a good practice to write all the functions including main() in the same class.

But as you opted for this, the code goes here like this:

public class CalcPyramidVolume {

/* Your solution goes here */

static double pyramidVolume(double baseLength, double baseWidth, double pyramidHeight)

{

double Volume,baseArea;

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public static void main (String [] args) {

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And there is not caveat of integer division, as you are declaring all your variables of type double.

3 0
4 years ago
Read 2 more answers
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Answer:

d

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<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

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5 0
3 years ago
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