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mojhsa [17]
3 years ago
6

What is a oxidation number

Physics
2 answers:
Scilla [17]3 years ago
6 0
Oxidation number is the charge.
Semmy [17]3 years ago
3 0
Oxodation numba is 2 i found out
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Which variables are involved in understanding Kepler's third law of motion? (1 point)
Zielflug [23.3K]

The variables which are involved in understanding Kepler's third law of

motion are

  • Orbital velocity
  • Distance to sun
<h3 /><h3>What is Kepler's third law of motion?</h3>

Kepler's third law of motion states that the the square of the orbital period of

a planet is proportional to the cube of the semi-major axis of its orbit. He

also inferred that the greater the distance, the slower the orbital velocity.

This thereby makes option D the most appropriate option as it contains the

orbital velocity and distance to sun variables.

Read more about Kepler's third law of motion here brainly.com/question/777046

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2 years ago
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Question 7<br> (01.01 MC)<br> What is the difference between a hypothesis and a theory?
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A transformer increases and decreases voltage.
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3 years ago
A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.
EleoNora [17]

(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done  in firing the projectile is 2,500 J.

<h3>Kinetic energy of the projectile at maximum height</h3>

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

  • m is mass of the projectile
  • v₀ₓ is the initial horizontal component of the velocity at maximum height

<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

<h3>Work done in firing the projectile</h3>

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

  • v is the resultant velocity

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done  in firing the projectile is 2,500 J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

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