__________ force is a fictitious force that some people believe causes you to feel as if you are being pushed outward from the c
2 answers:
You might be interested in
Answer:
b. Friction decreased when he went from pavement to ice and then increased two more times.
Explanation:
Frictional force depends on the normal force of the surface and a friction coefficient.
Since we're talking about the same car, the value of will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.
After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.
Answer:
F = 233.52 N, θ' = 351.41º
Explanation:
In this exercise we must find the net force applied on the donkey.
For this we use Newton's second law, where we create a reference frame with the horizontal x axis
let's decompose the forces
Jack
= 80.5 N
Jill
cos 45 = F_{2x} / F₂2
sin 45 = F_{2y} / F₂2
F_{2x} = F₂ cos 45
F_{2y} = F₂ sin 45
F_{2x} = 81.7 cos 45 = 57.77 N
F_{2y} = 81.7 sin 45 = 57.77 N
Jane
cos (270 + 45) = F_{3x} / F₃3
sin 315 = F_{3y} / F₃
F_{3x} = 131 cos 315 = 92.63 N
F_{3y} = 131 sin 315 = -92.63 N
the force can be found in each axis
X axis
F_{x} = F_{1x} + F_{2x} + F_{3x}
F_{x} = 80.5 +57.77 + 92.63
F_{x} = 230.9 N
Axis y
F_{y} = F_{1y} + F_{2y} + F_{3y}
F_{y} = 0 + 57.77 -92.63
F_{y} = -34.86 N
we can give the result in two ways
a) F = (230.9 i ^ - 34.86 j ^) N
b) in the form of module and angle
we use the Pythagorean theorem
F = √(Fₓ² + F_{y}²
F = √(230.9² + 34.86²)
F = 233.52 N
let's use trigonometry for the angle
tan θ =
θ = tan⁻¹ (\frac{F_y}{F_x} })
θ = tan⁻¹ (-34.86 / 230.9)
θ = -8.59º
if we measure this angle from the positive side of the x-axis counterclockwise
θ' = 360 -θ
θ‘= 360- 8.59
θ' = 351.41º