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natka813 [3]
3 years ago
10

Andy and Charlie are riding on a merry-go-round. Andy rides on a horse at the outer rim of the circular platform, twice as far f

rom the center of the circular platform as Charlie, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, which of the following best describes Andy's angular speed?'
Physics
1 answer:
Volgvan3 years ago
3 0

Complete Question:

Andy and Charlie are riding on a merry-go-round. Andy rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Charlie, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, which of the following best describes Andy's angular speed?'

( ) twice Charlie’s

( ) impossible to determine.

( ) the same as Charlie’s

( ) half of Charlie’s

Answer:

The same as Charlie's

Explanation:

As a merry-go-round is a rigid body, all points in the rotating body must have the same angular velocity, i.e., thet must rotate the same angle in the same time.

Otherwise, the distance between any pair of points on a given radius could be different in different times, which is not possible in a rigid body,

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A 0.250-kg aluminum bowl holding 0.800 kg of soup at 27.6°C is placed in a freezer. What is the final temperature if 424 kJ of e
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Answer:

16.32 °C

Explanation:

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Q = (m_b•c_b•Δt) + (m_s•c_s•Δt)

Where;

c_b = 0.215 kcal/(kg•°C)

c_s = 1 kcal/(kg•°C)

ΔT = 27.6 - 0 = 27.6°C

Thus;

Q = (0.25 × 0.215 × 27.6) + (0.8 × 1 × 27.6)

Q = 23.5635 Kcal

Now, the energy that exits to be used to freeze the soup is;

Q' = 424 kJ - Q

Let's convert 424 KJ to Kcal

424 KJ = 424/4.184 Kcal = 101.3384 Kcal

Thus;

Q' = 101.3384 - 23.5635

Q' = 77.7749 Kcal

Amount of heat that's removed is given by;

Q_f = Q' - mL

Where;

m = m_s = 0.8 kg

L = 79.8 kcal/kg

Thus;

Q_f = 77.7749 - (0.8 × 79.8)

Q_f = 13.9349 Kcal

Then final temperature will be;

T_f = Q_f/((m_b•c_b) + (m_s•c_s))

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T_f = 16.32 °C

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