If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that
<em>n</em> = 38 N
The friction force is proportional to the normal force by a factor of 0.27, so that
<em>f</em> = 0.27 (38 N) ≈ 10.3 N
and so the answer is D.
Answer:
vo=5.87m/s
Explanation:
Hello! In this problem we have a uniformly varied rectilinear movement.
Taking into account the data:
α =69.2
vf = 10m / s
h=2.7m
g=9.8m/s2
We know we want to know the speed on the y axis.
We calculate vfy
vfy = 10m / s * (sen69.2) = 9.35m / s
We can use the following equation.
We clear the vo (initial speed)
vo=5.87m/s
Answer:
149 m
Explanation:
The distances across the lake is forming a triangle.
let the distance between the point and the left side be 'x'
and the distance between the point and the right be 'y'
and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:
∠Z = 83°
x = 105 m
y = 119 m
Now, applying the Law of Cosines, we get
z² = x² + y² - 2xycos(Z)
Substituting the values in the above equation, we get
z² = 105² + 119² - 2×105×119×cos(83°)
or
z = √22140.48
or
z = 148.796 m ≈ 149 m
The point is 149 m across the lake
