a 0.500 kg mass is oscillating on a spring with k=330 N/m. the total energy of its oscillation is 3.24 J .what is the amplitude
2 answers:
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Answer:
<h2>A. 6pF</h2>Explanation:
If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown
Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;
C₄ ≈ 6pF
Hence the value of the X capacitor is approximately 6pF
Answer:
I = 2 kgm^2
Explanation:
In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:
(1)
I: moment of inertia of the door
α: angular acceleration of the door = 2.00 rad/s^2
τ: torque exerted on the door
You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:
(2)
F: force = 5.00 N
d: distance to the hinges = 0.800 m
You replace the equation (2) into the equation (1), and you solve for α:
Finally, you replace the values of all parameters in the previous equation for I:
The moment of inertia of the door around the hinges is 2 kgm^2
We know density = Mass / Volume
So Volume = Mass/Density
Volume = Area * Thickness
So the approximate thickness of the foil in millimeters =