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3 years ago

3 Alpha Centauri is approximately 4.3 light years

Physics

1 answer:

sergey [27]3 years ago

4 0

Answer:

40.68 trillion kilometers.

Explanation:

Multiply 4.3 × 9.46, the product which you get is your answer in trillion km

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The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

Andru [333]

(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

Fam

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3 years ago

What are the five basic postulates of kinetic-molecular theory?

lesantik [10]

Explanation:

The five basic postulate of kinetic molecular theory includes:

1) All gases consist of large amount and numbers of tiny particles that are far apart from each other and also relative to their size.

2) The collisions between gas particles and gas particles against container walls is refer to as elastic collision.

3) All gas particles are in a continuous random and rapid motion. They possess kinetic energy which is energy of motion.

4) There are no attractive force between gas particles.

5) The temperature of a gas depends on the average kinetic energy of the gas particle.

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3 years ago

We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav

Dmitrij [34]

Answer:

$t=1.5\times 10^{-4}\ s$

Explanation:

Given:

file size to be transmitted, $D=30\ Mb$

transmission rate of data, $\dot D=10\ Mb.s^{-1}$

propagation speed, $v=2\times 10^8\ m.s^{-1}$

distance of data transfer, $s=10000\ km=10^4\ m$

<u>Now the delay in data transfer from source to destination for each 10 Mb:</u>

$t'=\frac{s}{v}$

$t'=\frac{10^4}{2\times 10^8}$

$t'=5\times 10^{-5}\ s$

<u>Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:</u>

So,

$t=3\times t'$

$t=3\times 5\times 10^{-5}$

$t=1.5\times 10^{-4}\ s$

3 0

4 years ago

The magnitude R, measured on the Richter scale, of an earthquake of intensity I is defined as Requalslog StartFraction Upper I

lapo4ka [179]

Answer:

R = 6.8

Explanation:

Given data:

Richter scale $R = log(\frac{I}{I_o})$

where R - magnitude of earthquake of Richter scale

I - quake's intensity $= 10^{6.8} \times I_o$

$I_o$ - minimum intensity earthquake

Plugging all information in the equation to get Richter's scale

$R = log(\frac{10^{6.8} \times I_o}{I_o})$

$R = log(10^{6.8})$

R = 6.8

6 0

3 years ago

When antimatter interacts with an equal mass of ordinary matter, both matter and antimatter are converted completely into energy

Ilya [14]

Answer:

v = 5.88 10⁷ m / s

Explanation:

For this exercise we use the relation

E = m c²

also indicate that all energy is converted into kinetic energy

E = K = ½ (M-2m) v²

where m is the mass of antimatter and M is the mass of the ship's mass. Factor two is due to the fact that equal amounts of matter and antimatter must be combined

we substitute

m c² = ½ (M-2m) v²

v² = $2 \frac{m}{M+2m} \ c^2$

let's calculate

v = $\sqrt{2 \ \frac{4 \ 10^4 }{2 \ 10^6 + 2 \ 4 \ 10^4} \ (3 \ 10^8)^2}$

v = $\sqrt{ 34.615 \ 10^{14}}$

v = 5.88 10⁷ m / s

8 0

3 years ago