In 1976, the lockheed sr-71a blackbird set the record speed for any airplane: 3.53*10^3 km/h. suppose you observe this plane ascending at this speed. for 20.0 s, it flies at an angle of 15.0 degrees above the horizontal, then for another 10.0 s it's angle of ascent is increased to 35.0 degrees. calculate the plane's total gain in altitude, it's total horizontal displacement, and its resultant displacement.

v = speed of motion = 3530 km/h = 3530 (1000/3600) m/s = 980.6 m/s

consider the motion when the angle with the horizontal is 15 deg :

= component of velocity along vertical direction = v Sin15 = 980.6 Sin15 = 253.8 m/s

= component of velocity along horizontal direction = v Cos15 = 980.6 Sin15 = 947.2 m/s

t = time of travel = 20 sec

X = displacement along the x-direction

Y = displacement along the Y-direction

Horizontal displacement in first 20 sec is given as

X = t

X = 947.2 x 20

X = 18944 m

vertical displacement in first 20 sec is given as

Y = t

Y = 253.8 x 20

Y = 5076 m

consider the motion when the angle with the horizontal is 35 deg :

= component of velocity along vertical direction = v Sin35 = 980.6 Sin35 = 562.4 m/s

= component of velocity along horizontal direction = v Cos35 = 980.6 Sin35 = 803.3 m/s

t = time of travel = 10 sec

X' = displacement along the x-direction

Y' = displacement along the Y-direction

Horizontal displacement in next 10 sec is given as

X' = t

X' = 803.3 x 10

X' = 8033 m

vertical displacement in next 10 sec is given as

Y' = t

Y' = 562.4 x 10

Y' = 5624 m

X'' = total horizontal displacement

Y'' = total vertical displacement

Total horizontal displacement is given as

X'' = X + X' = 18944 + 8033 = 26977 m

Total vertical displacement is given as

Y'' = Y + Y' = 5076 + 5624 = 10700 m

Total gain in altitude = Y'' = 10700 m

total horizontal displacement = X'' = 26977 m

resultant displacement using pythagorean theorem is given as

R = sqrt(X''² + Y''²)

R = sqrt((26977)² + (10700)²)

R = 29021.5 m

  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