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MrRissso [65]
3 years ago
6

What are somethings about the Heliocentric model?

Physics
1 answer:
Rudik [331]3 years ago
6 0

Answer:

Heliocentrism, a cosmological model in which the Sun is assumed to lie at or near a central point (e.g., of the solar system or of the universe) while the Earth and other bodies revolve around it.

Explanation:

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What happens when red, blue, and green light come together?
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I believe if red, blue, and green light come together it would produce White light.

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1. The matter a wave travels through is called a
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It is called a medium
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What is the difference between torque and the moment of a force ​
STatiana [176]

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Torque Of a Force: If The Force has tendency or Bends The Body about Longitudinal axis of the Body it is Torque. Moment Of a Force :If Force has Tendency to or Rotates the Body about Transverse asis the Body It is Moment .

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2 years ago
If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links p
Mashcka [7]

The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²

By Newton's second law, the magnitude of the force <em>F</em> is proportional to the acceleration <em>a</em> according to

<em>F</em> = <em>m a</em>

where <em>m</em> is the object's mass. Solving for <em>m</em> gives

<em>m</em> = <em>F</em> / <em>a</em> = (10 N) / (3.52941 m/s²) ≈ 2.8 kg

4 0
2 years ago
Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s
g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

v_{x} =  v cos(∅)

v_{y} =  v sin(∅)

We know that for a projectile motion,

t =\frac{2vsin(∅)}{g}

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = v_{x}×t = vcos(∅)×\frac{2vsin(∅)}{g} = \frac{v^{2}sin(2∅) }{g}.

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = \frac{v^{2}sin(2α) }{g} = \frac{v^{2}sin(2[90-β]) }{g} =\frac{v^{2}sin(180-2β) }{g} = \frac{v^{2}sin(2β) }{g} .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

     β = 90-25 = 65°

∴ The required angle is 65°.

5 0
3 years ago
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