Does anyone know how to solve this?

110 m

Explanation:

Draw a free body diagram of the car.  The car has three forces acting on it: normal force up, weight down, and friction to the left.

Sum of the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of the forces in the x direction:

∑F = ma

-F = ma

-Nμ = ma

Substitute:

-mgμ = ma

-gμ = a

Given μ = 0.40:

a = -(9.8 m/s²) (0.40)

a = -3.92 m/s²

Given that v₀ = 30 m/s and v = 0 m/s:

v² = v₀² + 2aΔx

(0 m/s)² = (30 m/s)² + 2 (-3.9s m/s²) Δx

Δx ≈ 110 m

layer a is the exosphere and layer b is the thermosphere.   is the answer i think but i hope this .

  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