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Stolb23 [73]
1 year ago
15

After meeting the end the wave willa- reflect with larger amplitude.b- disappear.c- reflect with the same displacement. (not inv

erted)d- reflect with opposite displacement. (inverted)

Physics
1 answer:
Margarita [4]1 year ago
7 0

Answer:

c- reflect with the same displacement. (not inverted)

Explanation:

When a wave goes to an end that is free to move, the reflected wave will be equal to the initial wave but will go to the opposite direction. Since the end of the wave is free to move, the answer is

c- reflect with the same displacement. (not inverted)

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Which statement most accurately describes the second law of thermodynamics?
balandron [24]

Answer:

Energy cannot be changed from one form to another without a loss of usable energy

Explanation:

Second law of thermodynamics states that the total entropy or the randomness of the system remains constant over time.  It also states that the net entropy will remain the same or it will increase.

Entropy of a system is given by heat absorbed divided by temperature. It is given by :

\Delta S=\dfrac{\Delta Q}{T}

So, the correct option is (A) "Energy cannot be changed from one form to another without a loss of usable energy".

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3 years ago
One student did an experiment with two unknown minerals, Mineral 1 and Mineral 2. The hardness scale shown below was used for th
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Scale: (soft 1-->6 hardest)
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3 0
3 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:

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  • 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)

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Answer:

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