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2 years ago

7

Use the Electrostatic Series to predict the charges on the following pairs of substances when they are rubbed together:

1 answer:

Lisa [10]2 years ago

6 0

Answer: Not 100% sure but I believe the answer is B.

Hope this helps! ^^

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WILL GIVE BRAINLIEST PLEASE ANSWER ALL QUESTIONS

Rainbow [258]

There would be more - charges on the peanuts because they gain electrons which are represented with the - sign. There would also be more + on the cat to show that the cat’s fur is positively charged. The peanuts would be attracted to the cat because opposite charges attract.

As for explaining the first model, there should be the same amount of + and - charges on the peanuts as well as the cat to show that it is balanced and the peanuts are not attracted or repelled by the cat.

5 0

2 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

Two tuning forks are played at the same time. One has a frequency of 490 Hz and the other is 488 Hz. How many beats per second a

Gnoma [55]

Hope this helps you.

7 0

3 years ago

Read 2 more answers

não é difícil de perceber uma relação aditiva entre os termos consecutivos (1;3;6;10...)(A)qual é o sexto e décimo número triang

EastWind [94]

It is 15 and 45 ............

7 0

3 years ago

What force is needed to accelerate a child on a sled (total mass = 65.0 kg ) at 1.25 m/s2?

ratelena [41]

Given; Mass m = 65.0 Kg; acceleration a = 1.25 m/s²

Required: Force F = ?

Formula: F =ma

F = (65.0 Kg)(1.25 m/s²)

F = 81.25 N

4 0

2 years ago