Plzzzzzzz help me 15 pt for 1st 10 for 2nd
2 answers:
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Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;
(a) the force constant of the spring
(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k
Answer:
Explanation:
Firstly, when you measure the voltage across the battery, you get the emf,
E = 13.0 V
In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.
Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (). But this is not the case.
Let the resistance of the ammeter be r
Hence, using Ohm's law we get the following 2 equations:
.......(1)
......(2)
Substituting the value of r from (2) in (1), we have,
which simplifying gives us, (which is our required solution)
putting the value of z in either (1) or (2) gives us, r = 0.5325