What will the stopping distance be for a 3,000-kg car if -3,000 n of force are applied when the car is traveling 10 m/s? a. 30 m b. 150,000 m c. 50 m d. 10 m

F = force applied to stop the car = - 3000 N

m = mass of the car = 3000 kg

a = acceleration of the car = ?

v₀ = initial velocity of the car before the force is applied to stop it = 10 m/s

v = final velocity of the car when it comes to stop = 0 m/s

d = stopping distance of the car

acceleration of the car is given as

a = F/m

inserting the values

a = - 3000/3000

a = - 1 m/s²

using the kinematics equation

v² = v²₀ + 2 a d

inserting the values

0² = 10² + 2 (-1) d

0 = 100 - 2 d

2 d = 100

d = 100/2

d = 50 m

hence the correct choice is

C. 50 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

f=ma newton

f=100x2.5=250n