The electric force between two charge objects is calculated through the Coulomb's law.
F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m.
Both don’t have units beacuse they are ratios
We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is
Where,
F_D = Drag Force
= Drag coefficient
A = Area
= Density
V = Velocity
Our values are given by,
(That is proper of a cone-shape)
Part A ) Replacing our values,
Part B ) To find the torque we apply the equation as follow,
Answer:
612000 C
Explanation:
Current, I, is given as the rate of flow of charge, that is:
I = Δq / Δt
where q = electric charge
t = time taken
This implies that:
Δq = I * Δt
The battery rating is 170 Ampere-hours, therefore:
Δq = 170 * 1 hour
But 1 hour = 3600 seconds;
=> Δq = 170 * 3600 = 612000 C
The total charge that the battery can provide is 612000 C.
Answer:
W = F * s
Work done equals applied force * distance traveled
Apparent weight = M g (1 - sin θ) since some of applied force will lighten sled
μ = coefficient of kinetic friction
F cos θ = force applied to motion of sled
s = distance traveled
[μ M g (1 - sin θ)] cos θ * s = work done in moving sled
Note that F = μ M g if applied force is in the horizontal direction
