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adelina 88 [10]
3 years ago
11

airground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow ha

s an 11 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration 2.3 times that due to gravity
Physics
1 answer:
aliina [53]3 years ago
4 0

Answer:

N = 13.65 rpm

Explanation:

given data

radius = 11 m

centripetal acceleration =  2.3 times that due to gravity

to find out

how many revolutions per minute

solution

we know here centripetal accel = 2.3 × g

ω²r  = 2.3 × 9.8

ω² × 11  = 2.3 × 9.8  

solve it we get

ω² = 2.0490

ω = 1.43 rad/s

and

ω = \frac{2\pi N}{60}

1.43 = \frac{2\pi N}{60}  

solve it we get

N = 13.65 rpm

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Hint: Here, we can see that 12 and 13 are consecutive numbers. So, all numbers between squares of 12 and 13 are non-square numbers. Therefore, first find squares of 12 and 13 and then subtract square of 12 from square of 13, we get numbers of non-square numbers. At the last subtract 1 from the result obtained as both extremes numbers are not included.

Complete step-by-step answer:

In these types of questions, a simple concept of numbers should be known that is between squares of two consecutive numbers all numbers are non-square numbers. Also one tricky point should remember that whenever we find the difference between two numbers we get a number of numbers between them including anyone of the extreme numbers. So we subtract 1 to exclude both extreme numbers.

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