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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 62kg box is lifted 12 meters off the ground. How much work is done?
Temka [501]

Answer: 7291.2 joules

Explanation:

Work is done when force is applied on an object over a distance.

Thus, Workdone = Force X distance

Since Distance moved by box = 12 metres

mass of box = 62kg

Acceleration due to gravity when box was lifted is represented by g = 9.8m/s^2

Recall that Force = Mass x acceleration due to gravity

i.e Force = 62kg x 9.8m/s^2

= 607.6 Newton

So, Workdone = Force X Distance

Workdone = 607.6 Newton X 12 metres

Workdone = 7291.2 joules

Thus, 7291.2 joules of work was done.

4 0
3 years ago
Which could be an example of a balanced force?
Rus_ich [418]

Answer:

An  asteroid moving at a constant speed through space.

Explanation:

4 0
3 years ago
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A train increases its speed steadily from 10 m/s to 20 m/s in
rewona [7]

Answer:

15m/s

Explanation:

add the two speeds and divide by 2

10+20=30

30/2=15

3 0
3 years ago
Please I really need help on these
blagie [28]

Answer:

  3, 4, 2

Explanation:

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4 0
3 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0
2 years ago
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