Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º
The direction of an electric field is determined from the behavior of a positive test charge that is set free in the electric field.This charge moves along a distinct vector showing the direction of the electric field Therefore the answer is b. a positive charge will move in the field.
The strength of the fireman in vertical direction will be given by F = m * g. Then, the work done will be given by definition by W = F * d. Substituting the expression of the Force in that of the work, we have that the work will be W = m * g * d. Substituting the given values and assuming that g = 10m / s ^ 2, we have a total work of W = (73) * (10) * (9) = 6570 J
Answer:
Hindi ko alam pasensya ka ha godbless
Answer:
Plug in the given values and solve for the final velocity. Remember, when the ball is on the ground it has a height of zero.
Explanation:
