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3 years ago

When will the solar system die

Physics

2 answers:

scoundrel [369]3 years ago

8 0

No the system will never doe that's just like people say in the world is going to end

Alenkinab [10]3 years ago

3 0

I would say our solar sustem will die when the sun dies or a black hole sucks it up.

You might be interested in

Write with all the steps and formulas and drawing if needed.

lianna [129]

<u>The answer is not contained detail explanation, just a solution and the required values. </u>

All the details are in the pictures, the answers are marked with orange colour.

Note,

in the task no 20.:

$m_A - the \ mass \ of \ A; \ m_B-the \ mass \ of \ B \ balls.\\V_A \ and \ V_B-the \ velocities \ of \ the \ A&B \ balls \ before \ collision.\\V'_A \ and \ V'_B-the \ velocities \ of \ the \ A&B \ balls \ after \ collision.$

V - the velocity of the pair of the balls after collision.

in the task no 21:

m₁ - the mass of the copper ball; m₂ - the mass of the copper calorimeter; m₃ - the mass of the water; t₀ - the initial temperature of water in the copper calorimeter; θ - the final temperature in the calorimeter after the copper ball is transferred into a copper calorimeter; t₁ - the required initial temperature of the copper ball before it is transferred into the calorimeter.

7 0

3 years ago

A material kept at high temperature is seen to emit photons with energies of 0.3 eV, 0.5 eV, 0.8 eV, 2.0 eV, 2.5 eV, and 2.8 eV.

sukhopar [10]

Answer:

0.3 eV, 0.5eV,, 8 eV, 2.0eV, 2.50 eV, 2.8 eV

Explanation:

In a given material the emission and absorption spectra are equivalent, for which the emission spectrum observed at high temperature for the material corresponds to the transition between the energy states of the material, the process is that the electrons exist from the ground state until an excited state and after a short period of time or these electrons relax emitting photons.

In the absorption process, the material is at low temperature, ideally at A = 0K, whereby all states are in the ground state and all excited states are empty. therefore it can absorb the beam energy for each transition given from the ground state to each excited edtado.

Consequently, the lines above the absorption oscillate lines coincide with the lines of emotion, this we see lines oscillate at 0.3 eV, 0.5eV,, 8 eV, 2.0eV, 2.50 eV, 2.8 eV

5 0

3 years ago

A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 an

Tom [10]

Answer:

0.69s

Explanation:

10 cm = 0.1 m

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

$\omega = \alpha t = 2.1 t$

And so the radial acceleration is

$a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2$

The tangential acceleration is always the same since angular acceleration is constant:

$a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2$

For these 2 quantities to be the same

$a_r = a_t$

$0.441 t^2 = 0.21$

$t^2 = 0.21/0.441 = 0.4762$

$t = \sqrt{0.4762} = 0.69 s$

6 0

4 years ago

Which two statements about composite materials are true?

Lostsunrise [7]

Answer:

Here down below is 3 things true about composite materials

Explanation:

-They’re made up of more than one substance.

-They have the same or similar properties as the materials used to make them.

-They’re always made of metal.

-They’re readily available in nature.

4 0

3 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Scrat [10]

Answer:

5.24 m/s

Explanation:

So for the rod to be able to rise upward to the straight up position, the kinetic energy caused by linear speed v0 must be just enough to convert into the potential energy.

Since the rod is uniform in mass, we can treat the body as 1 point, at its center of mass, or geometric center, aka 0.35 / 2 = 0.175 m from the pivot.

For the rod to swing from bottom to top, the center must have moved a distance of h = 0.175 * 2 = 0.35 m, vertically speaking.

Since we neglect friction and air resistance, according to the law of energy conservation then:

$E_k = E_p$

$mv^2/2 = mgh$

Where v is the speed at the center of mass, g = 9.81 m/s2 is the gravitational acceleration, and m is the mass. We can divide both sides by m

$v^2 = 2gh = 2*9.81*0.35 = 6.867$

$v = \sqrt{6.867} = 2.62 m/s$

As this is only the speed at the center of mass, the speed at the bottom end would be different, to calculate this, we need to find the common angular speed:

$\omega = v / r = 2.62 / 0.175 = 14.97 rad/s$

Where r is the rotation radius, or the distance from pivot point to the center of mass

$v_0 = \omega R = 14.97*0.35 = 5.24 m/s$

Where R is the distance from the pivot to the bottom end of the rod

5 0

3 years ago