An oscillating block-spring system has a mechanical energy of 4.72 J, an amplitude of 11.2 cm, and a maximum speed of 4.24 m/s.
1 answer:
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Answer:
Centripetal acceleration = 83.77m/s²
Explanation:
<u>Given the following data;</u>
Radius, r = 0.13m
Velocity, v = 3.3m/s
To find centripetal acceleration;
Centripetal acceleration is given by the formula;
Substituting into the equation, we have;
<em>Centripetal acceleration = 83.77m/s²</em>
<em>Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s². </em>
Answer:
F = 3.6 kN, direction is 9.6º to the North - East
Explanation:
The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.
Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.
Wind
X axis
F₁ = 2.50 kN
Tide
cos 30 = F₂ₓ / F₂
sin 30 = F_{2y} / F₂
F₂ₓ = F₂ cos 30
F_{2y} = F₂ sin 30
F₂ₓ = 1.20cos 30 = 1.039 kN
F_{2y} = 1.20 sin 30 = 0.600 kN
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = 2.50 +1.039
Fₓ = 3,539 kN
F_y = F_{2y}
F_y = 0.600
to find the vector we use the Pythagorean theorem
F =
F =
F = 3,589 kN
the address is
tan θ = F_y / Fₓ
θ = tan⁻¹
θ = tan⁻¹ 0.6 / 3.539
θ = 9.6º
the resultant force to two significant figures is
F = 3.6 kN
the direction is 9.6º to the North - East