The restoring force of the spring cancels the weight of the mass, so by Newton's second law
∑ F = F[spring] - mg = 0 ⇒ F[spring] ≈ 45.1 N
where m = 4.60 kg and g = 9.80 m/s². Then the spring constant is k such that by Hooke's law,
F[spring] = k x
where x = 0.0231 m. Then the spring constant is
k = F[spring]/x ≈ 1950 N/m
Answer:
0.98 g/m
Explanation:
Note: Since Tension and frequency are constant,
Applying,
F₁²M₁ = F₂²M₂............... Equation 1
Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.
make M₂ the subject of the equation
M₂ = F₁²M₁/F₂²............... Equation 2
From the question,
Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz
Substitute these values into equation 2
M₂ = 196²(0.31)/110²
M₂ = 0.98 g/m
Explanation:
The angle of the handle relative to the horizontal is 35°. The angle of the ramp to the horizontal is 7°. So the angle of the handle relative to the ramp is 28°.
cos 28° = 50 / F
F = 50 / cos 28°
F = 56.6 lbs
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