Acommercial passenger jet is flying with an airspeed of 185 miles per hour on a heading of 036°. if a 47-mile-per-hour wind is blowing from a true heading of 120°, determine the velocity and direction of the jet relative to the ground.

195.5 mph, 49.85 degree

Explanation:

velocity of jet with respect to wind = 185 mph at 36 degree

V (j, w) = 185 (Cos 36 i + Sin 36 j) = 149.67 i + 108.74 j

velocity of wind with respect to ground = 47 mph at 120 degree

V(w,g) = 47 (Cos 120 i + Sin 120 j) = - 23.5 i + 40.70 j

Velocity of jet with respect to wind = velocity of jet with respect to ground - velocity of wind with respect to ground

V (j, w) = V (j, g) - V (w, g)

149.67 i + 108.74 j = V(j,g) + 23.6 i - 40.7 j

V(j,g) = 126.07 i + 149.44 j

Magnitude of velocity of jet with respect to wind

= = 195.5 mph

Let it makes angle θ from + X axis

tan θ = 149.44 / 126.07 = 1.1185

θ = 49.85 degree

  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