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guapka [62]
3 years ago
5

• How were historical solar system models different from our current model?

Physics
1 answer:
horrorfan [7]3 years ago
4 0
One of the most famous historical model was the geocentric model thought of by plato. This model says that the earth is in the center of the universe and that each of the planets follow complicated paths that go backwards sometimes. This model was used until a new one was accepted. This model is the one we have today called the heliocentric model. It was bought of by Copernicus. Hope that answers your question!
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A 10 Kg ball is rolling at 2.5 m/s. It is then hit from behind with a bat that puts a 300 N force on the ball for a quick .3 sec
vazorg [7]

Answer: g. gg g rfrcdv

Explanation:

vfv g bygyb

7 0
2 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
2 years ago
What does the word “adverse” mean in the following sentence? Adverse reactions, such as fever and headache, can occur if the med
Stells [14]
I think it's D. unfavorable hope it helps
3 0
3 years ago
Read 2 more answers
What is likely happening at the boundary between the Nazca plate and the Pacific plate?
Nataly_w [17]
<span>The eastern margin is a convergent boundary subduction zone under the South American Plate and the Andes Mountains, forming the Peru–Chile Trench. The southern side is a divergent boundary with the Antarctic Plate, the Chile Rise, where seafloor spreading permits magma to rise.</span>
8 0
3 years ago
Read 2 more answers
A student of mass 51 kg wants to walkbeyond the edge of a cliff on a heavy beam ofmass 220 kg and length 11 m. The beam isnot at
pishuonlain [190]

Answer:

4.465 m

Explanation:

Taken the beam to be uniform, then the center of gravity will act at the mid-point of the beam. The mid point = 11 / 2 = 5.5 m

since the boy can walk to the end of the beam without falling,

then the torque ( moment of the force) by beam = torque by students at the end of the beam

let the distance of the boy = x

220 ( 5.5 - x ) = 51 × x

1210 - 220 x = 51 x

1210 = 51 x + 220 x

1210 = 271 x

x = 1210 / 271 = 4.465 m

3 0
3 years ago
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