Atoy rocket is launched straight up by using a spring. the rocket is initially pressed down on the spring so that the spring is compressed by 9 cm. if the spring constant is 830 n/m and the mass of the rocket is 50 g, how high will the rocket go? you may neglect the effects of air resistance. a 14.48 m b 6.86 m c 10.84 m d 3.91 m e 8.03 m

B.6.86 m

Explanation:

Given that

Compression in spring ,x= 9 cm

Spring constant ,K= 830 N/m

Mass of the rocket ,m= 50 g = 0.05 kg   ( 1 kg = 1000 g)

Now by applying energy conservation

Now by putting the values ( we are taking g= 9.81 m/s²)

h= 6.853 m

Therefore the answer is B.

B.6.86 m

  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

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.