4. The value of the before and after-collision momentum of two colliding objects is shown in the

A series L-R-C circuit consists of a 226 Ω resistor, a 27.4 mH inductor, a 11.55 µF capacitor, and an AC source of amplitude 15

DanielleElmas [232]

Answer: 363 Ω.

Explanation:

In a series AC circuit excited by a sinusoidal voltage source, the magnitude of the impedance is found to be as follows:

Z = √((R^2 )+〖(XL-XC)〗^2) (1)

In order to find the values for the inductive and capacitive reactances, as they depend on the frequency, we need first to find the voltage source frequency.

We are told that it has been set to 5.6 times the resonance frequency.

At resonance, the inductive and capacitive reactances are equal each other in magnitude, so from this relationship, we can find out the resonance frequency fo as follows:

fo = 1/2π√LC = 286 Hz

So, we find f to be as follows:

f = 1,600 Hz

Replacing in the value of XL and Xc in (1), we can find the magnitude of the impedance Z at this frequency, as follows:

Z = 363 Ω

6 0

3 years ago

SO! I'VE GOT A COUPLE QUESTIONS RELATING TO SCIENCE (WHERE ARE MY SCIENCE WHIZZES?) TRUST ME THERE WILL BE MORE

Helga [31]

Answer:

nnnnnnnnnnnnnnnnnnnnnnnn

Explanation:

5 0

3 years ago

Read 2 more answers

All of the outer planets are much larger than the inner planets true or false

Rama09 [41]

That statement is false because yes Jupiter and Saturn are large, however, planets like Uranus and Neptune are quite small if not smaller than some of the inner planets.

4 0

3 years ago

Which of the following was not a factor that got the public interested in NASA? A. real people B. scientific reports C. adventur

fomenos

B. Scientific reports

8 0

3 years ago

Read 2 more answers

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago