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ArbitrLikvidat [17]

2 years ago

14

Which device is similar to a sandwich that has layers of semiconductors?

2 answers:

lawyer [7]2 years ago

8 0

Answer:

b

Explanation:

DENIUS [597]2 years ago

5 0

Answer:

C. A transistor

Explanation:

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Find the acceleration if a 32.5 N force is<br> used on an object that has a mass of<br> 128.6 kg.

Sholpan [36]

Answer:

Acceleration=3.95

Explanation:Use the formula a=m/f

a=128.6/32.5

a=3.95

3 0

2 years ago

1. Set frequency of wave generator to 5Hz.

ivann1987 [24]

Separate the barriers so they have a 2cm gap between them.

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3 years ago

I really need help on this question!

steposvetlana [31]

Answer:

option "c"

Explanation:

because in gases molecules are further apart and move very quickly

4 0

3 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago

The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

motikmotik

<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = $\frac{Gm_1m_2}{r^2}$

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = $\sqrt{\frac{Gm_1m_2}{F} }$

If we substitute the values in the above equation

r = $\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }$

= 2.46 m

3 0

3 years ago

Read 2 more answers