Answer: C
Frictional force
Explanation:
The description of the question above is an example of a circular motion.
For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.
Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.
Therefore, the correct answer is option C - the frictional force.
Answer:
Re = 1 10⁴
Explanation:
Reynolds number is
Re = ρ v D /μ
The units of each term are
ρ = [kg / m³]
v = [m / s]
D = [m]
μ = [Pa s]
The pressure
Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]
μ = [Pa s] = [kg / m s²] [s] = [kg / m s]
We substitute the units in the equation
Re = [kg / m³] [m / s] [m] / [kg / m s]
Re = [kg / m s] / [m s / kg]
RE = [ ]
Reynolds number is a scalar
Let's evaluate for the given point
Where the data for methane are:
viscosity μ = 11.2 10⁻⁶ Pa s
the density ρ = 0.656 kg / m³
D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m
Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶
Re = 1.19 10⁴
