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mestny [16]
3 years ago
10

What are the symmetry operations of molecule AB4, where the molecule b lies at the center of the square and A lies at the center

of the square and is not coplaner with B atoms. How to find the multiplication table
Physics
1 answer:
arlik [135]3 years ago
4 0

The symmetry operations of molecule AB₄ is tetrahedral  and they are :

  • E : Identity
  • 4C₃ : axis of rotation
  • 3C₂ : axis of rotation
  • 3S₄ : Rotation-reflection axis
  • 6бd  

To find the multiplication table we have to apply the multiplication table for Td symmetry ( attached below )

Given that molecule B lies at the corners of the square and molecule A lie at the center and is not coplanar with molecule B the symmetry operations of the molecules AB₄ will belong to a tetrahedral symmetry group which contains :

  • E : Identity
  • 4C₃ : axis of rotation
  • 3C₂ : axis of rotation
  • 3S₄ : Rotation-reflection axis
  • 6бd

Learn more about tetrahedral symmetry group : brainly.com/question/1968705

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The Force on the left hand pole, F' = 0.167N

<h3>What is the force on the left hand pole?</h3>

Force is an agent which produces a change in the motion or state of an object.

Force is a vector quantity.

The general force is calculated as follows:

F = mg/sinθ

m = 17.1 g = 0.0171 kg

g = 9.81 m/s²

θ = 45°

F = 0.0171 * 9.81/sin45

F = 0.237 N

Force on the left hand pole, F' = Fcosθ

F' = 0.237 * cos 45

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In conclusion, the force on the left hand pole is the horizontal component of force.

Learn more about force at: brainly.com/question/141439

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An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it
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Answer:

μs = 0.36

Explanation:

  • While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.
  • This acceleration is produced by the centripetal force.
  • Now, this force is not a different type of force, is the net force acting on the riders in this direction.
  • Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.
  • In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
  • When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:

       F_{frmax} = \mu_{s}* F_{n}  = m * g  (1)

  • where μs= coefficient of static friction (our unknown)
  • As  we have already said Fn = Fc.
  • The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:

      F_{n} = m* \omega^{2} * r  (2)

  • Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:

       \mu_{s} = \frac{g}{\omega^{2} r}  (3)

  • Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:

       \omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)

  • Replacing g, ω and r in (3):
  • \mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)

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