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3 years ago

The solar nebula was a

Physics

1 answer:

tangare [24]3 years ago

4 0

Solar Nebula

Our solar system began forming within a concentration of interstellar dust and hydrogen gas called a molecular cloud. The cloud contracted under its own gravity and our proto-Sun formed in the hot dense center. The remainder of the cloud formed a swirling disk called of the solar nebula.

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Copper has a specific heat of 0.386 J/g°C. How much heat is required to increase 5.00 g of copper from 0.0°C to 10.0°C?

Leto [7]

The answer is 19.3 j

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2 years ago

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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

2 years ago

g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

sdas [7]

Answer:

The vertical distance is $d = \frac{2}{k} *[mg + f]$

Explanation:

From the question we are told that

The mass of the cylinder is m

The kinetic frictional force is f

Generally from the work energy theorem

$E = P + W_f$

Here E the the energy of the spring which is increasing and this is mathematically represented as

$E = \frac{1}{2} * k * d^2$

Here k is the spring constant

P is the potential energy of the cylinder which is mathematically represented as

$P = mgd$

And

$W_f$ is the workdone by friction which is mathematically represented as

$W_f = f * d$

$\frac{1}{2} * k * d^2 = mgd + f * d$

=> $\frac{1}{2} * k * d^2 = d[mg + f ]$

=> $\frac{1}{2} * k * d = [mg + f ]$

=> $d = \frac{2}{k} *[mg + f]$

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3 years ago

Which of the following statements best summarizes the main points of the passage?

Deffense [45]

Answer:the summery is ...i really don't know the picture is blurry and I cant see can you make it clear?

Explanation:

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1 year ago

A bike travels at a constant speed for 4.00 m/s for 5.00 seconds. How far does it go

Umnica [9.8K]

Answer:

20 m

Explanation:

4.00 m/s x 5 seconds

4 m/s x 5 s = 20m

7 0

3 years ago

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