Physics, 25.01.2021 20:20, jaliyahmw

The maximum displacement in an oscillatory motion is A = 0.49 m. Determine the position x at which the kinetic energy of the particle is half its elastic potential energy. (If KE = U/2 x= ?).

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answered: trint4

x = 0.40 m

Explanation:

When the displacement is maximum, the particle is momentarily at rest, which means that at this point (assuming no friction present) all the mechanical energy is elastic potential, which can be written as follows:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} (1)$

Since in absence of friction, total mechanical energy must keep constant, this means that at any time, the sum of the kinetic and potential energy, must be equal to (1), as follows:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} = (KE)_{f} + U_{f} (2)$

If KEf = U/2f, replacing in (2), we get:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} = (U/2)_{f} + U_{f} = \frac{3}{2} *U_{f} (3)$

At any point, the elastic potential energy is given by the following expression:

$U_{f} = \frac{1}{2} *k*x^{2} (4)$

where k= spring constant (N/m) and x is the displacement from the

equilibrium position.

Replacing (4) in (3), simplifying and rearranging, we get:

$E_{tot} = U_{o} = \frac{1}{2} *A^{2} = \frac{3}{4} *x^{2} (5)$

Finally, solving for x, we get:

$x = \sqrt{\frac{2}{3} } * A = \sqrt{\frac{2}{3} } * 0.49m = 0.40 m (6)$

answered: Guest

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

answered: Guest

second mirror gathers light from a faraway object and reflects it to another mirror. - a.

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The maximum displacement in an oscillatory motion is A = 0.49 m. Determine the position x at which t...