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hoa [83]
3 years ago
15

Which of the following statements about planetary satellites is true? all planetary satellites are as large as our moon or bigge

r. all planetary satellites are relatively small, typically having a radius of no more than 20 kilometers. planetary satellites vary greatly in size, but none are larger than any of the planets in the solar system. planetary satellites vary greatly in size, from very small, to some that are larger than some planets.
Physics
2 answers:
Illusion [34]3 years ago
6 0
<span> planetary satellites vary greatly in size, from very small, to some that are larger than some planets.</span>
Anvisha [2.4K]3 years ago
3 0

Answer:

Planetary satellites vary greatly in size, from very small, to some that are larger than some planets.

Explanation:

Natural sateletes are celestial bodies that orbit other larger celestial bodies. The solar system has several satellites, the best known being the moon, the natural satellite of planet Earth. The planets of the solar system have different sizes, just like the natural satellites. Thus, there are satellites larger than planets. The moon itself is a larger satellite than some planets, like Mercury.

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In an intergalactic competition, spaceship pilots compete to see who can cover the distance between two asteroids in the short-
pogonyaev

Answer:

a)  truc is C,  b) correct result is the B

Explanation:

As the speed of the competition is very high, for the judges the speed is

           v = d / t

           v = 3 109 m / 20

           v = 1.5 108 m / s

This is half the speed of light. For these high speeds we must use the relations of special relativity.

For the time          t = to γ

For distance         L = Lo / γ

                            γ = √ (1-v2 / c2)

Own time and distance (to and Lo) corresponds to the observer who is not moving the judges in this case

Let's look for the range value

                     γ = 1 / √ (1 - (1.5 / 3) 2) = 1 / 0.866 = 1.15

The time              t = 20 1.15 = 23 s

The distance       L = 3 10 9 /1.15 = 2.60 109 m

From these results we see that time increases and the distance is shorter.

Let's review the claims

A) False. It's the opposite

B) False

C) True. It is according to the result found

D) False.

In the nuclear fusion process, we will also use the special relativity that has a relationship between energy and mass

         ΔE = c² Δm

As in the process energy is released, for the law of conservation of the mass of energy to be fulfilled, the total mass of the products, He atom, must be reduced.

Therefore the correct result is the B

4 0
3 years ago
To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newt
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Answer: A.E.

Explanation: To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newton's law of gravitation. The concepts of mass and weight are often confused. In fact, in everyday conversations, the word "weight" often replaces "mass," as in "My weight is seventy-five kilograms"

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3 years ago
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A wind turbine turns wind energy into electricity using the aerodynamic force
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Answer:

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6 0
2 years ago
The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the
tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

<em>A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.</em>

<em>A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.</em>

<em>B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]</em>

<em />

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0
3 years ago
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