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exis [7]
3 years ago
9

The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which

can function as a second hand (so that it makes one revolution every 60 s).
Part A: Find the speed of the passengers when the Ferris wheel is rotating at this rate (in m/s)
Part B: A passenger weighs 834 N at the weight-guessing booth on the ground. What is his apparent weight at the highest point of the Ferris wheel? (in N)
Part C: What is his apparent weight at the lowest point on the Ferris wheel?
Part D: What would be the time for one revolution if the passeger's apparent weight at the highest point were zero? (in s)
Part E: What then would be the passenger's apparent weight at the lowest point? (in N)
Physics
1 answer:
emmainna [20.7K]3 years ago
8 0

Answer:

Explanation:

A ) angular velocity ω = 2π / T

= 2 x 3.14 / 60

= .10467 rad / s

linear velocity v = ω R

=  .10467 x 50

= 5.23 m / s

centripetal force = m v² / R

= mg v² / gR

= 834 x 5.23² / 9.8 x 50

= 46.55 N

B )

apparent weight

= mg - centripetal force

= 834 - 46.55

= 787.45 N

C ) apparent weight

= mg + centripetal force

= 834 + 46.55

= 880.55 N.

D )

For apparent weight to be zero

centripetal force = mg

mg = mv² / R

v² = gR

= 9.8 x 50

= 490

v = 22.13 m /s

time period of revolution

= 2π R /v

2 x 3.14 x 50 / 22.13

= 14.19 s  

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accelerated motion

Explanation:

a change in velocity (10 m/s to 50 m/s) over time (5 s) is called acceleration.

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Which conditions are usually the effect of a low air pressure system?
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Explanation : High air pressure is generally associated with nice weather.

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3 years ago
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A spelunker is surveying a cave. She follows a passage that takes her a distance 184 m straight west, then a distance 220 m in a
Sever21 [200]
Refer to the diagram shown below.

Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.

The first distance is 184 m west. It is represented by
d₁ = -184 i

The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j

The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) =  102.42 i + 18.06 j

Let the fourth distance be 
d₄ = a i + b j

Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.

The x-component yields
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a = -108.95

The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94

The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m

The direction is at an angle θ north of west, given by
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Answer:
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Answer:

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Explanation:

Given that

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Mass of oxygen ,m₂ = 32 u

u₂  = 401 m/s

v₂ =445 m/s

Given that initially both are moving in the same direction and lets take they are moving in the right direction.

Speed of the helium after collision = v₁

There is no any external force on the masses that is why the linear momentum will be conserve.

Initial linear momentum = Final linear momentum

P = m v

m₁u₁+m₂u₂ = m₁v₁+m₂v₂

598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445

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