Answer:
1) true
2) false
3) false
4) true
5) true
6) true
7) true
8) false
9) true
10) false
i think these are correct if im wrong on a few im sorry. Hope this helps at least a bit. And if i do get some wrong you know just to pick the opposite answer.
Answer:
The particles will more likely to move faster since they are converted from a liquid to gas.
Rules for States of Matter:
1. Solid particles always are packed close together and don't have much space to move.
2. Liquid particles have space to move around but are still packed together, but not as close as solid.
3. Gas particles are moving freely, in fact they are in the air! Gas particles are free to move wherever. For example, the air has gas particles that are constantly bumping into each other.
Let me know if I am right =)
<em>The correct option is </em><em>A</em>. The information we know about the known exoplanets is estimates of orbits and masses.
<h3>What is exoplanets?</h3>
An exoplanet or extrasolar planet is a planet outside the Solar System.
In other words, exoplanet is any planet beyond our solar system.
<h3>Characteristics of exoplanets</h3>
exoplanets are known for the following characteristics;
- they are usually hot
- they can orbit their stars so tightly that a “year” lasts only a few days
- they can orbit two suns at once
Thus, the information we know about the known exoplanets is estimates of orbits and masses.
Learn more about exoplanets here: brainly.com/question/1514493
#SPJ1
Gravitational potential energy :)
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)
q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
