We know that the Moon revolves around Earth during a period of 27.3 days. The average distance from the center of Earth to the center of the Moon is 3.84×108 m.

1)What is the magnitude of the acceleration of the Moon due to its motion around Earth?

Explanation:

This is a circular motion questions

Where the oscillation is 27.3days

Given radius (r)=3.84×10^8m

Circular motion formulas

V=wr

a=v^2/r

w=θ/t

Now, the moon makes one complete oscillation for 27.3days

Then, one complete oscillation is 2πrad

Therefore, θ=2πrad

Then 27.3 days to secs

1day=24hrs

1hrs=3600sec

Therefore, 1day=24×3600secs

Now, 27.3days= 27.3×24×3600=2358720secs

t=2358720secs

Now,

w=θ/t

w=2π/2358720 rad/secs

Now,

V=wr

V=2π/2358720 ×3.84×10^8

V=1022.9m/s

Then,

a=v^2/r

a=1022.9^2/×3.84×10^8

a=0.0027m/s^2

  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

c) up is the direction the magnetic force is acting on the charge.