"A 0.4 kg mass, attached to the end of a 0.75 m string, is whirled around in a circular horizontal path. If the maximum tension that the string can withstand is 450 N, then what maximum velocity can the mass have if the string is not to break?"

The maximum velocity the mass can have if the string is not to break = 29.05 m/s

Explanation:

The force balance in the mass:

The tension in the string must always be equal to the force keeping the mass in horizontal circular motion.

The force keeping the mass in circular motion is given by

F = mv²/r

m = mass of body = 0.4 kg

v = speed of the body in circular motion

r = radius of the circular motion = 0.75 m

Maximum tension the string can withstand will correspond to the maximum velocity of the body in horizontal circular motion

T = F = mv²/r

450 = (0.4)(v²)/(0.75)

v² = 450×0.75/0.4 = 843.75

v = 29.05 m/s

Explanation:

Below is an attachment containing the solution.

  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