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Delvig [45]
3 years ago
15

You sit at the outer rim of a Ferris Wheel that rotates 2 revolutions per minute (RPM). What would your rotational speed be if y

ou were instead clinging to a position halfway from the center to the outer rim. Be sure to provide at least 3 to 4 complete content related sentences and show any work needed to support your answer
Physics
2 answers:
svetoff [14.1K]3 years ago
5 0
Your rotational speed would still be the same. This is because all parts of the Ferris wheel rotate together. Your linear speed however would change. That is a function of radius. But the question is asking about rotational speed and that does not change in this situation
AURORKA [14]3 years ago
5 0

Answer:

The rotational speed will be the same

Explanation:

When there is a circular motion involved (as in this case with the Ferris Wheel), the rotational speed is defined as the ratio between the angle covered \Delta \theta and the time taken to cover that angle (\Delta t):

\omega = \frac{\Delta \theta}{\Delta t}

Since all the points along a radius of the the Ferris Wheel rotates coherently, they cover the same angular distance in the same amount of time, so the rotational speed is always the same for every point of the Wheel. This means that if you change the distance by moving halfway from the centre, the rotational speed is still 2 revolutions per minutes, because a point halfway from the centre also covers 2 complete revolutions in a minute.

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Answer:

Apply the following formulae horizontally And get A value for time

Remember horizontal acceleration is zero

s  = ut +  \frac{1}{2}a {t}^{2}   \\ 0.8 = 1.7 \times t \\  \frac{0.8}{1.7}  = t \\ t = 0.47s

and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

s = ut +  \frac{1}{2} a {t}^{2}  \\ s =  \frac{1}{2}  \times 10 \times (0.47) ^{2}  \\ s = 1.1045m

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2 years ago
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choli [55]

Answer:

The average net force on the truck is 375 Newtons.

Explanation:

Using Newton's 3rd equation of motion, we have :

v^{2} - u^{2} = 2×a×s

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u = initial velocity = 20 m/s

a = acceleration

s = distance traveled = 300 m

Using these values in the above equation, we get acceleration = 0.375 m/s^{2}

Using Newton's second law, we have:

F=m×a

where m = mass = 1000 kg

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Putting values we have F=375 N

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Answer:

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6 0
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