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4 years ago

What causes interstellar dust and clouds into plants and stars

1 answer:

5 0

I assume you mean planets eitherway interstellar dust and clouds have solid material and as well as minerals such as iron and over time solid chuncks start to fuse together and make a planet but with stars however it is different because a star has a little solid core surrounded by gasses (hydrogen and helium) to produce heat in a method called nuclear fusion so interstellar clouds contain gasses as well and solids as I previously said and form a star as well

this is from my understanding

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A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char

Mila [183]

Answer:

Ek = 8,79 [J]

Explanation:

We are going to solve this problem, using the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁ = Ek + U₁

As charge are at rest Ek = 0

And U₁ has two components

U₁₂ = K * Q₁*Q₂ / 0,4 and U₃₂ = K*Q₃*Q₂ / 0,6

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4 ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6 ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J] ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then E₁ = E₁₂ + E₃₂

E₁ = 10,5 [J]

At the moment of Q₂ passing x = 40 cm or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂ = U₁₂ + U₃₂

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7 ⇒ U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7 U₁₂ = 7,71 [J]

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3 ⇒ U₃₂ = - 9*20*10*10⁻³/0,3

U₃₂ = - 9*20*10⁻²/0,3

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as

E₂ = Ek + U₂ and E₂ = E₁

Then

Ek + U₂ = E₁

Ek = 10,5 - U₂

Ek = 10,5 - 1,71

Ek = 8,79 [J]

5 0

3 years ago

John often sees squiggly lines when he looks at a bright sky or white paper. His doctor says that they are called eye floaters,

trapecia [35]

Answer- It's B, i double checked it on a few websites

8 0

3 years ago

Read 2 more answers

A bicycle of mass m requires 50 J of work to move from rest to a final speed v. If the same amount of work is performed during t

insens350 [35]

Answer:

The final speed of the second bicycle is (v·√2)/2

Explanation:

The mass of the given bicycle = m

The amount of work required to move the bicycle from rest to speed v = 50 J

The final speed of the first bicycle = v

The mass of the second bicycle = 2m

Therefore, from conservation of energy, we have;

Work required by the first bicycle = Kinetic energy gained by the bicycle

The kinetic energy = 1/2·m·v²

∴ Energy required by the first bicycle = 50 J = 1/2·m·v²

Given that the same amount of work is performed on the second bicycle, we have;

Work performed on the second bicycle = 50 J = kinetic energy of second bicycle = 1/2·(2·m)·v₂²

Also, given that 50 J = 1/2·m·v², we have;

Work performed on the second bicycle = 50 J = 1/2·m·v²= 1/2·(2·m)·v₂²

1/2·m·v²= 1/2·(2·m)·v₂²

m·v² = 2·m·v₂²

v² = 2·v₂²

v₂ = √(v²/2) = v/√2 = (v·√2)/2

v₂ = (v·√2)/2

The final speed of the second bicycle = v₂ = (v·√2)/2.

3 0

3 years ago

Based on Newton's law of motion, which combination of rocket bodies and engine will result in the acceleration of 40 m/s ^2 at t

ad-work [718]

The question is incomplete. The complete question is :

The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.

A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.

Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?

Body 3 + Engine 1

Body 2 + Engine 2

Body 1 + Engine 2

Body 1 + Engine 1

Solution :

Given :

Body Mass (gram) Engine Force (newtons)

1 500 1 25

2 1500 2 20

3 750 3 30

The body 1 has a mass of 500 gram which is equal to 0.5 kg

And engine 2 has a force of 20 newtons.

We know that according to Newton's laws of motion,

Force = mass x acceleration

20 = 0.5 x acceleration

Acceleration $=\frac{20}{0.5}$

$=\frac{200}{5}$

$= 40 \ m/s^2$

Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of $40 \ m/s^2$ at the start of the launch.

8 0

3 years ago

A car has a mass of 1300 kg and a velocity of 15 m/s. The car crashes into a wall and stops in two seconds. What is the stopping

Musya8 [376]

Answer:

Explanation:

Newton's second law:

F = ma

The acceleration is Δv/Δt, so:

F = m Δv/Δt

Given m = 1300 kg, Δv = 15 m/s, and Δt = 2 s:

F = (1300 kg) (15 m/s) / (2 s)

F = 9750 N

Answer B.

3 0

3 years ago