A car of mass 1270 kg makes a 15.0 m radius turn at 11.8 m/s on flat ground. How much centripetal force acts on it?
1 answer:
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Answer:
154° at 195 km/h
Explanation:
The helicopter is moving south at 175 km/h, relative to the wind.
But the wind is moving east at 85 km/h, relative to the ground.
This means that the helicopter is moving south east relative to the ground.
Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.
This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.
Refer to the triangle b1.
The distance traveled by the helicopter in 1 hour is denoted by d.
d is the hypotenuse of the right triangle.
Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)
Hence the helicopter is traveling at 195 km/h relative to the ground.
To calculate the direction we use,
tan (x) = opposite/adjacent = 85/175
So the angle x is,
= 25.9°
Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)
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