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den301095 [7]
3 years ago
11

Consider a series RLC circuit where R = 855 Ω and C = 6.25 μF. However, the inductance L of the inductor is unknown. To find its

value, you decide to perform some simple measurements. You apply an ac voltage that peaks at 84.0 V and observe, using an oscilloscope, that the resonance angular frequency occurs at 51900 s–1 (recall that ω = 2π f). What is the inductance of the inductor in millihenrys?
Physics
1 answer:
sashaice [31]3 years ago
6 0

Answer:

L= 0.059 mH

Explanation:

Given that

R = 855 Ω and C = 6.25 μF

V= 84 V

Frequency

ω = 51900 1/s

We know that

\omega=\sqrt{\dfrac{1}{LC}}

L=Inductance

C=Capacitance

ω =angular Frequency

ω² L C =1

(51900)² x L x 6.25 x 10⁻⁶ = 1

L= 5.99 x 10⁻⁵ H

L= 0.059 mH

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Question 1 (1 point)
puteri [66]

Answer:

250 N

Explanation:

a =  \frac{vf - vi}{t}

Vf=final velocity

Vi =initial velocity

=70-20/2=25m/s^2

F=ma

=10kg * 25m/s^2

=250N

3 0
3 years ago
What is the magnitude of vector K ?
vovangra [49]

Answer:

Vector K = 8 m

Explanation:

The given figure shows a right angle triangle JKL. It is given that :

vector L = 10 m

vector J = 6 m

We have to find vector k. We can find it by using Pythagoras theorem. According to this theorem, the sum of squares of perpendicular and the base is equal to the square of the longest side.

L^2=K^2+J^2

K^2=L^2-J^2

K^2=(100)^2-(6)^2

K = 8 m

Hence, the magnitude of vector K is 8 m.

3 0
3 years ago
One piece of copper jewelry at 111°C has exactly twice the mass of another piece, which is at 28°C. Both pieces are placed insid
vladimir2022 [97]

Answer:

83.33 C

Explanation:

T1 = 111 C, m1 = 2m

T2 = 28 C, m2 = m

c  = 0.387 J/gK

Let the final temperature inside the calorimeter of T.

Use the principle of calorimetery

heat lost by hot body = heat gained by cold body

m1 x c x (T1 - T) = m2 x c x (T - T2)

2m x c X (111 - T) = m x c x (T - 28)

2 (111 - T) = (T - 28)

222 - 2T = T - 28

3T = 250

T = 83.33 C

Thus, the final temperature inside calorimeter is 83.33 C.

3 0
3 years ago
At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6
Elis [28]

The particle has constant acceleration according to

\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k

Its velocity at time t is

\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du

\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t

\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k

Then the particle has position at time t according to

\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du

\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k

At at the point (3, 6, 9), i.e. when t=0, it has speed 8, so that

\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64

We know that at some time t=T, the particle is at the point (5, 2, 7), which tells us

\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}

and in particular we see that

v_{0y}=-2v_{0x}

and

v_{0z}=-v_{0x}

Then

{v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3

\implies v_{0y}=\mp\dfrac{8\sqrt6}3

\implies v_{0z}=\mp\dfrac{4\sqrt6}3

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k

4 0
3 years ago
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