The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:
(1)
The work done is equal to the elastic energy stored by the compressed spring:
where
is the spring constant and
is the compression of the spring. If we substitute the numbers, we find:
And now we can use eq.(1) to calculate the average power output:
Answer:
(a) I_A=1/12ML²
(b) I_B=1/3ML²
Explanation:
We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².
(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².
(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:
Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:
Finally, using the Parallel Axis Theorem, we calculate I_B:
Answer:
100 cm³
Explanation:
Use ideal gas law:
PV = nRT
where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.
n and R are constant, so:
P₁V₁/T₁ = P₂V₂/T₂
If we say point 1 is at 40m depth and point 2 is at the surface:
P₂ = 1.013×10⁵ Pa
T₂ = 20°C + 273.15 = 293.15 K
P₁ = ρgh + P₂
P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa
P₁ = 4.933×10⁵ Pa
T₁ = 4.0°C + 273.15 = 277.15 K
V₁ = 20 cm³
Plugging in:
(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)
V₂ = 103 cm³
Rounding to 1 sig-fig, the bubble's volume at the surface is 100 cm³.
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sorry this is by accident, but i can't delete it
