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Kaylis [27]
3 years ago
8

The tallest Ferris wheel in the world is located in Singapore. Standing 42 stories high and holding as many as 780 passengers, t

he Ferris wheel has a diameter of 150m and takes approximately 30 minutes to make a full circle. Determine the speed of the riders (in m/s) on the Singapore Flyer.
Physics
1 answer:
kakasveta [241]3 years ago
8 0

Answer:

The speed of the riders on the Singapore Flyer is approximately 0.262 m/s

Explanation:

The dimensions of the tallest Ferris wheel in the world are;

The diameter of the Ferris wheel, D = 150 m

The tine it takes the Ferris wheel to make a full circle, T = 30 minutes = 30 min × 60 s/min = 1,800 seconds

The angular velocity of the Ferris wheel, ω = 2·π/T

The linear velocity of the Ferris wheel, v = r·ω = The speed of the riders

Where;

r = The radius of the Ferris wheel = D/2

D = 150 m

∴ r = 150 m/2 = 75 m

∴ v = r·2·π/T

∴ v = 75 m × 2 × π/(1,800 s) ≈ 0.262 m/s

The speed of the riders on the Singapore Flyer, v ≈ 0.262 m/s

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3 years ago
An object with little mass won't require a lot of force to move.
Anna35 [415]

Answer:

True

Explanation:

Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects. Example: Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop.

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3 years ago
A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s
pashok25 [27]

Answer:

109656.25 Nm

Explanation:

\omega_f = Final angular velocity = 1.5 rad/s

\omega_i = Initial angular velocity = 0

\alpha = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm

The torque specifications must be 109656.25 Nm

5 0
3 years ago
A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?
Varvara68 [4.7K]

Answer:

<h2>15.25 N</h2>

Explanation:

       A force of 55\text{ }N is acting on a wagon along the road. The wagon weights 7.5\text{ }kg. Acceleration of the wagon is given as 5.3\text{ }\frac{m}{s^{2}}.

       Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

       Gravitational Force and Normal Reaction cancel out each other.

       Net External Force = Mass of system/wagon \times Acceleration of wagon

       F_{ext}-F_{friction}=(7.5\text{ }kg)\times(5.3\text{ }\frac{m}{s^{2}})=39.75\text{ }N\\55\text{ }N-F_{friction}=39.75\text{ }N\\F_{friction}=15.25\text{ }N

F_{friction} has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N

4 0
3 years ago
Can anyone heelp me plzz
lina2011 [118]

Answer:

the ans is D... good luck

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