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polet [3.4K]
3 years ago
15

The Cosmo Clock 21 Ferris wheel in Yokohama, 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 60.0 s).
(a) Find the speed of the passengers when the Ferris wheel is rotating at this rate.
(b) A passenger weighs 882 N at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel?
(c) What would be the time for one revolution if the passenger’s apparent weight at the highest point were zero?
(d) What then would be the passenger’s apparent weight at the lowest point?
Physics
1 answer:
Sonja [21]3 years ago
6 0

Answer:

Explanation:

a ) speed of passenger = circumference / time

= 2π R / Time

= 2 x 3.14 x 50 / 60

= 5.23 m /s

b )

centrifugal force = m v² /R

= (882 /9.8 ) x 5.23² / 50

= 77.47 N

Apparent weight at the highest point

real weight  - centrifugal force

= 882 - 77.47

= 804.53 N

Apparent weight at the lowest point

real weight  + centrifugal force

= 882 +77.47

= 959.47 N

c )  if the passenger’s apparent weight at the highest point were zero

centrifugal force = weight

mv² /R = mg

v² = gR

= 9.8 X 50

v = 22.13 m /s

d )

apparent weight

mg - mv² / R

= 882 - (882 / 9.8 )x 22.13²/50

= 882 +  882

= 1764 N  

=

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Which of the following shows the units of angular motion?
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4 0
3 years ago
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0
3 years ago
Heather and Jerry are standing on a bridge 49 mm above a river. Heather throws a rock straight down with a speed of 17 m/sm/s .
lara [203]

Answer:

3.467 s

Explanation:

given,

distance , d = 49 mm = 0.049 m

initial speed of the of the rock, v = 17 m/s

time taken by the Heather rock to reach water

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s = ut +\dfrac{1}{2}at^2

taking downward as negative

-0.049 = -17 t -\dfrac{1}{2}\times 9.8\times t^2

4.9 t² + 17 t - 0.049 = 0

now,

t_1 = \dfrac{-(17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}

t₁ = -3.47 s , 0.0028 s

rejecting negative values

t₁ = 0.0028 s

now, time taken by the ball of Jerry

using equation of motion

s = ut +\dfrac{1}{2}at^2

taking downward as negative

-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2

4.9 t² - 17 t - 0.049 = 0

now,

t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}

t₂ = 3.47 s ,-0.0028 s

rejecting negative values

t₂ = 3.47 s

now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s

5 0
3 years ago
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