The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s−1)t]. (a) at time
1 answer:
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Answer:
72.98 km
Explanation:
Her displacement is simply the distance from her final position to her initial position.
Now, I've drawn and attached a triangle diagram to depict this her movement.
Point O is her initial starting point.
Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.
From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.
Thus;
x² = 62² + 26² - 2(62 × 26)cos 120
x² = 4520 + 806
x² = 5326
x = √5326
x = 72.98 km
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:3.95 m/s
Explanation:
Given
mass of object
radius of circle
initial Position
angular displacement
8.95 radian can be written as
i.e. Particle is at first quadrant with
(c)velocity is