A ball is thrown with velocity 30.m⋅s−1at an angle of 40.º below the horizontal. What is the magnitude of the horizontal component of the velocity? Give your answer to the nearest m⋅s−1 and do not include units with your answer.

Explanation:

The horizontal component of the velocity is expresssed as;

Vx = Vcos θ

V is the velocity of the ball

θ is the angle of inclination to the horizontal

Given

V = 30m/s

θ = 40°

Vx = 30cos40°

Vx = 22.98

Hence the magnitude of the horizontal component of the velocity to the nearest m/s is 23

answer: its c lololo

  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