Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F =
q =
we calculate
q =
q = Ra 7 10-12
q = 2.65 10⁻⁶ C
Answer:
F = 3.6 kN, direction is 9.6º to the North - East
Explanation:
The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.
Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.
Wind
X axis
F₁ = 2.50 kN
Tide
cos 30 = F₂ₓ / F₂
sin 30 = F_{2y} / F₂
F₂ₓ = F₂ cos 30
F_{2y} = F₂ sin 30
F₂ₓ = 1.20cos 30 = 1.039 kN
F_{2y} = 1.20 sin 30 = 0.600 kN
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = 2.50 +1.039
Fₓ = 3,539 kN
F_y = F_{2y}
F_y = 0.600
to find the vector we use the Pythagorean theorem
F =
F =
F = 3,589 kN
the address is
tan θ = F_y / Fₓ
θ = tan⁻¹
θ = tan⁻¹ 0.6 / 3.539
θ = 9.6º
the resultant force to two significant figures is
F = 3.6 kN
the direction is 9.6º to the North - East
Answer: The trip takes
Explanation:
Velocity is the variation of the position of a body (distance traveled
) with time
:
In this case, the car travels a distance at a velocity
and we need to find the time it takes the trip.
Isolating :
Finally: