Physics, 17.07.2019 02:10 katherinelondon4002

In experiment 1, when you measured the output with the 51 ohm resistor load in place, you probably still noticed some loading effect. this is because an op amp can only output a certain amount of current before it begins behaving nonlinearly. every op amp has a maximum positive current, called "sourcing maximum negative current, called "sinking current" that can flow into the output terminal of the op amp.) look up the lf412 spec sheet and find nominal value(s) for this sourcing current limit. based on this value, determine the minimum load resistance that could be driven by the lf412 op amp without experiencing any loading effects in experiment 1 3.

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Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

a skateboard rolling across the street - apex

Answer from: Quest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

Answer from: Quest

At maximum height, kinetic energy is converted in potential energy formula: at max height, ke = ½ * 0.27kg * (16m/s) ² = 34.56 j so the remaining pe = (41.70 – 34.56)= 7.14 j = 0.27 kg * 9.8m/s² * h = (9.8/7.14 * 0.27)so max height h = 5.08 or 5 m

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In experiment 1, when you measured the output with the 51 ohm resistor load in place, you probably s...