L. An arrow is shot at 35° above the horizontal. Its velocity is 52 m/s and it hits the target. a. What is the horizontal velocity? (Show Work!)

The horizontal component of the velocity is approximately 42.6 m/s

Explanation:

The given parameters are;

The speed of the arrow when it was shot = 52 m/s

The angular direction in which the arrow was shot, θ = 35° above the horizontal

The horizontal component of the velocity is found by the resolution if the velocity vector into its components as follows;

The horizontal component of the velocity = Velocity in the x-direction = vₓ

vₓ = v × cos(θ) = 52 × cos (35°) ≈ 42.6 m/s

vₓ = The horizontal component of the velocity ≈ 42.6 m/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