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fredd [130]
3 years ago
10

In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b

acks against the wall. the cylinder begins to rotate about a vertical axis. then the floor on which the passengers are standing suddenly drops away! if all goes well, the passengers will "stick" to the wall and not slide. clothing has a static coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. what is the minimum rotational frequency, in rpm, for which the ride is safe?
Physics
1 answer:
Vesna [10]3 years ago
7 0
<span>The friction force ( µ . N ) must be enough to counter the gravity force ( m . g ) The Normal force from the rotating wall is the centripetal force = mv^2 / r so with the lowest coefficient of friction ( 0.4 ) 0.4 . v^2 / 2.5 = g v^2 = 2.5 . 9.8 / 0.4 v = 7.83 m/s T = 5π / 7.83 = 2.007 s f = 0.5 Hz = 30 rpm</span>
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d) 12 V

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A grandfather clock is controlled by a swinging brass pendulum that is 1.2 m long at a temperature of 27°C. (a) What is the leng
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Answer:

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

Explanation:

Given;

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Final temperature T2 = 0.0°C

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The change i length ∆L;

∆L = L2 - L1

L2 = L1 + ∆L ...........1

∆L = xL1(∆T)

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Substituting the given values into equation 2;

∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)

∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)

∆L = -6.156 × 10^-4 m

From equation 1;

L2 = L1 + ∆L

Substituting the values;

L2 = 1.2 m + (- 6.156 × 10^-4 m)

L2 = 1.2 m - 6.156 × 10^-4 m

L2 = 1.1993844 m

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

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