Arocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s but does not stop. m/s2 how high does it rise above the ground?

consider the motion of rocket until it runs out of fuel

v₀ = initial velocity = 0 m/s

v = final velocity when it runs out of fuel = ?

t = time after which fuel is finished = 3.98 sec

a = acceleration = 29.4 m/s²

Y₀ = height gained when the fuel is finished = ?

using the kinematics equation

v = v₀ + a t

v = 0 + (29.4) (3.98)

v = 117.01 m/s

using the equation

v² = v²₀ + 2 a Y₀

(117.01)² = 0² + 2 (29.4) Y₀

Y₀ = 232.85 m

consider the motion of rocket after fuel is finished till it reach the maximum height.

Y₀ = initial position = 232.85 m

Y = final position at maximum height

v₀ = initial velocity just after the fuel is finished = 117.01 m/s

v = final velocity after it reach the maximum height = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

using the kinematics equation

v² = v²₀ + 2 a (Y - Y₀)

inserting the values

0² = (117.01)² + 2 (- 9.8) (Y - 232.85)

Y = 931.4 m

33: i think it’s b

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂