Answer:
If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.
Explanation:
To determine how fast the car was moving, after skidding, the formula below is used:
V = √32*fd
V is the car's speed (ft/s)
d is skid length (ft) = 66 ft
f is the coefficient of friction determined by the material the car was skidding on.
Coefficient of friction for wet concrete is 0.65
V = √32*fd
V = √32 *0.65* 66
V = 242.679 ft/s ≅ 243 ft/s (nearest whole number)
If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.
The correct answer is an ''electric current''.
Answer: 0.43 V
Explanation:
L = [μ(0) * N² * A] / l
Where
L = Inductance of the solenoid
N = the number of turns in the solenoid
A = cross sectional area of the solenoid
l = length of the solenoid
7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24
1.752*10^-3 = 4π*10^-7 * 202500 * A
1.752*10^-3 = 0.255 * A
A = 1.752*10^-3 / 0.255
A = 0.00687 m²
A = 6.87*10^-3 m²
emf = -N(ΔΦ/Δt).........1
L = N(ΔΦ/ΔI) so that,
N*ΔΦ = ΔI*L
Substituting this in eqn 1, we have
emf = - ΔI*L / Δt
emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3
emf = 0.0234 / 0.055
emf = 0.43 V
I believe the answer is C- payload