A900-kg car initially at rest rolls 50 m down a hill inclined at an angle of 5.0°. a 400-n effective friction force opposes its motion.
a. what distance will it travel on a similar horizontal surface (assume the friction force is the same) at the bottom of the hill?
b. if the friction force remains the same will the distance decrease or increase if the car's mass is 1800 kg?
o will decrease
o will increase
o will remain the same

d = 146.2 m

increase

Explanation:

Using Energy Balance:

U_i + W =  U_f

K_i + U_gi + W = K_f + U_fg

0 + m*g*h + F*d = 0.5*m*v_f^2

(900)*(9.81)*(50 sin (5)) + 400*50 = 0.5*900*v_f^2

v_f^2 = 129.944

v_f = 11.4 m/s at the bottom of the hill

U_i + W =  U_f

K_i + U_gi + W = K_f + U_fg

0.5*m*v_i^2 - F*d = 0

0.5*900*129.944 = 400*d

d = 146.19 m

part b

Heavier car will gain greater momentum at the bottom of the hill and attain a higher velocity at the bottom; hence, the distance will increase.

increase

  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