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nataly862011 [7]
3 years ago
7

A 75.0 kg person is on a Ferris Wheel that has a radius R = 16 m. The tangential velocity of the Ferris Wheel is 8.25 m/s. Calcu

late the period of rotation of the Ferris Wheel. Enter your answer below and include units and round to 3 significant figures. Show your work on paper, scan it, and submit it in the last problem.

Physics
1 answer:
Dovator [93]3 years ago
3 0

Answer:

The period of rotation is

T=8.025s

Explanation:

The person is undergoing simple harmonic motion on the wheel

Given data

mass of the person =75kg

Radius of wheel r=16m

Velocity =8.25m/s

The oscillating period of simple harmonic motion is given as

T=(2*pi)/2=2*pi √r/g

Assuming that g=9.81m/s

Substituting our data into the expression we have

T=2*3.142 √ 16/9.81

T=6.284*1.277

T=8.025s

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Two resistors, A and B, are connected in a series circuit with a battery. The resistance of A is twice that of B. Which resistor
IgorC [24]

Answer:

A dissipates more power.

Explanation:

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How do H-R diagrams help scientists compare new stars with the Sun? In your answer, name one type of data that H-R diagrams do n
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Answer

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Explanation:

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When reading a buret, where is the initial and final volumes taken from? The top (where the zero is) or the bottom?. If the bure
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3 years ago
An ice cream maker has a refrigeration unit which can remove heat at 120 Js'. Liquid ice
Rom4ik [11]

Answer:

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C is 45,360 J

Explanation:

The given parameters for the refrigeration unit and the ice cream are;

The power of the refrigeration unit = 120 J/s

The mass of the liquid ice cream, m = 0.6 kg

The initial temperature of the liquid ice cream, T₁ = 20°C

The freezing point temperature of the ice cream, T₂ = -16°C

The specific heat capacity of the ice cream, c = 2,100 J/kg⁻¹·°C⁻¹

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, ΔQ, is given as follows;

ΔQ = m × c × ΔT

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Therefore, by substituting the known values, we have;

ΔQ = 0.6 × 2,100 × (20 - (-16)) = 45,360

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