Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the drawstring back 60 cm from its rest position and holds it in position with a force of 166 N. If the mass of the arrow is 54 g and the "spring" is massless, what is the speed (in m/s) of the arrow immediately after it leaves the bow?

42.95m/s

Explanation:

The magnitude of the force of the spring is given as

kx = 166N

From the question, we know that x = 60cm = 0.6m

Then, the constant of the spring would be k = 166/0.6 = 276.7N/m

When the bow is pulled back 60cm, the potential energy of the arrow is.

PE = 1/2kx²

PE = 0.5 * 276.67 * 0.6²

PE = 49.8J

Finally, the speed of the arrow can be found by using principle of conservation of energy, that states that, the total energy of an isolated system remains constant and conserved over time. This means, energy is neither created nor destroyed.

1/2kx² = 1/2mv²

kx² = mv²

v² = kx²/m

v² = 276.67 * 0.6²/ 0.054

v² = 1844.5

v = 42.95m/s

  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

seeing the graph would