Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with the radius of RE). Use the gravitational acceleration on the surface of the Earth is 9.8m/s2.

The magnitude of the free-fall acceleration at the orbit of the Moon is (, where ).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration (), in meters per square second, is directly proportional to the mass of the Earth (), in kilograms, and inversely proportional to the distance from the center of the Earth (), in meters:

(1)

Where:

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

- Mass of the Earth, in kilograms.

- Distance from the center of the Earth, in meters.

If we know that , and , then the free-fall acceleration at the orbit of the Moon is:

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

their crystal patterns have been disrupted by stress