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Alex
3 years ago
6

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

rotates around its axis. Once the Rotor reaches its operating speed, the floor drops but the riders remain pinned against the wall of the cylinder. Suppose the cylinder makes 26.0 rev/min and has a radius of 3.70 m. 1) What is the coefficient of static friction between the wall of the cylinder and the backs of the riders
Physics
1 answer:
steposvetlana [31]3 years ago
3 0

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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3 years ago
In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin
Natasha2012 [34]

Answer:

y = 0.0233 m

Explanation:

In a Young's Double Slit Experiment the distance between two consecutive bright fringes is given by the formula:

Δx = λL/d

where,

Δx = distance between fringes

λ = wavelength of light

L = Distance between screen and slits

d = Slit Separation

Now, for initial case:

λ = 425 nm = 4.25  x 10⁻⁷ m

y = 2Δx = 0.0177 m => Δx = 8.85 x 10⁻³ m

Therefore,

8.85 x 10⁻³ m = (4.25 x 10⁻⁷ m)L/d

L/d = (8.85 x 10⁻³ m)/(4.25 x 10⁻⁷ m)

L/d = 2.08 x 10⁴

using this for λ = 560 nm = 5.6 x 10⁻⁷ m:

Δx = (5.6 x 10⁻⁷ m)(2.08 x 10⁴)

Δx = 0.0116 m

and,

y = 2Δx

y = (2)(0.0116 m)

<u>y = 0.0233 m</u>

3 0
3 years ago
A beam of white light enters a prism of crown glass (n = 1.5) from air (n = 1.00). Once inside, the colors in the light disperse
eduard

Answer:

= 2.33

Explanation:

.According to snell's law:

n1sin i = n2sin r ,

where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,

i is angle of incidence,

r is angle of refraction.

Given that,

n1 = 1,

i = 51 degrees,

r = 19.5 degrees. ,

n2= ?

So,

1*sin 51 = n2 sin 19.5  

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3 years ago
This illustration shows two opposing forces pulling on a wagon. Which description best describes how the wagon will move?
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Answer:

The wagon will move to the right.

Explanation:

From the question given above, the following data were obtained:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Direction of the wagon =.?

To determine the direction in which the wagon will move, we shall determine the net force acting on the wagon. This can be obtained as follow:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 30 – 10

Fₙ = 20 N to the right

From the calculations made above, the net force acting on the wagon is 20 N to the right. Hence the wagon will move to the right.

8 0
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You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.
dem82 [27]
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  • = (20 × 200 × 1/2) J
  • = (20 × 100) J
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<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0
2 years ago
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