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valkas [14]
3 years ago
9

Date:

Physics
1 answer:
tia_tia [17]3 years ago
4 0

Answer:

don't know I just want point for answers

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Imagine you want to create a model of Solar system (and beyond) with the Sun
kondaur [170]

Using the scale model of the Sun given;

  • The diameter of the Milky Way = 4.28 × 10¹¹ m
  • The diameter of the Earth = 5.44 × 10⁻³ m

<h3>What is the diameter of the Milky Way?</h3>

The diameter of the Milky Way is about 1 × 10¹⁸ km.

The diameter of the Sun is about 1.4 × 10⁶ km

The diameter of the Earth is about 1.27 × 10⁴ km.

Using the scale model of the Sun given, the diameter of the Milky Way = (1 × 10¹⁸ km/1.4 × 10⁶ km) × 0.6 m

The diameter of the Milky Way = 4.28 × 10¹¹ m

Using the scale model of the Sun given, the diameter of the Milky Way = (1.27 × 10⁴ km/1.4 × 10⁶ km) × 0.6 m

The diameter of the Earth = 5.44 × 10⁻³ m

In conclusion, the diameter of the Milky Way is far bigger than the Sun while the diameter of the Sun is about 5400 times bigger than the Earth.

Learn more about the Milky Way, Sun, and Earth at: brainly.com/question/1995133

#SPJ1

5 0
2 years ago
A balloon holds 50.0 mL of nitrogen at 25° C and a pressure of 736 mm Hg.
olganol [36]

Answer:

Explanation:

4 0
3 years ago
GIVING BRAINLIEST PLEASE HELP!!
levacccp [35]

Answer: wood

Explanation: pls help me brainly.com/question/22777864

7 0
3 years ago
Which of the following is not an intensive physical property?
Alex Ar [27]

Answer;

-Mass

Explanation;

-An intensive property is a physical quantity whose value does not depend on the amount of the substance for which it is measured. For example, boiling point, density, color, melting point, Odor, temperature, etc.

-Extensive properties on the other hand, do depend on the amount of matter that is present. An extensive property is considered additive for subsystems. Examples of extensive properties include: volume, mass, size, weight. length, etc.


5 0
4 years ago
Read 2 more answers
An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. at time t = 0.00
Masteriza [31]
The cart is moving by simple harmonic motion, and its position at time t is described by
x(t) = A \cos (\omega t)
where
A is the amplitude of the oscillation
\omega is the angular frequency

The amplitude of the oscillation corresponds to the maximum displacement of the spring, which corresponds to the initial position where the spring was released: 
A=0.250 m

The period of the motion is T=5.67 s, and the angular frequency is related to the period by
\omega =  \frac{2 \pi}{T}= \frac{2 \pi}{5.67 s} =1.11 rad/s

Therefore now we can calculate the position of the system at the time t=29.6 s:
x(29.6 s)=(0.250 m)\cos ((1.11 rad/s)(29.6 s))=+0.033 m
3 0
3 years ago
Read 2 more answers
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