Consider a car with front cross-sectional area of A=(1.77±0.02)m^2 and a drag coefficient of C=(0.38±-.02) moving with a speed of v=(10.0±0.5)m/s. Suppose that the density of air is p=1.20 kg/m^3 at the sea level.
Calculate the force F=(1/2)CApv^2 on the car due to air friction.
To calculate the force on the car due to air friction, you need to use the formula F = (1/2)CApv^2, where F is the force, A is the front cross-sectional area, C is the drag coefficient, p is the density of air, and v is the velocity of the car.
Given values:
A = (1.77 ± 0.02) m^2 (front cross-sectional area of the car)
C = (0.38 ± 0.02) (drag coefficient)
p = 1.20 kg/m^3 (density of air at sea level)
v = (10.0 ± 0.5) m/s (velocity of the car)
To calculate the force, we need to take into account the uncertainties in the measurements. We will use the method of propagation of uncertainties to account for this.
Step 1: Calculate the product (1/2)CAp.
a) Calculate the central value:
(1/2)CAp = (1/2) * (1.77 ± 0.02) * (0.38 ± 0.02) * (1.20 ± 0) = (1/2) * 1.77 * 0.38 * 1.20
b) Calculate the uncertainties:
Δ[(1/2)CAp] = |(1/2)CAp| * √[(ΔA/A)^2 + (ΔC/C)^2 + (Δp/p)^2]
Step 2: Calculate the force F.
a) Calculate the central value:
F = (1/2)CApv^2 = (1/2)CAp * (10.0 ± 0.5)^2
b) Calculate the uncertainties:
ΔF = |F| * √[(Δ[(1/2)CAp]/[(1/2)CAp])^2 + (Δv/v)^2]
Plug in the values and uncertainties into the formulas and calculate the force F.