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REY [17]
4 years ago
7

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle. At the bottom of the dive, the speed of

the airplane is 320 m/s. What is the smallest radius allowable for the vertical circle if the pilot's apparent weight is not to exceed 7.0 times his true weight
Physics
1 answer:
crimeas [40]4 years ago
4 0

Answer:

1.74 km

Explanation:

We are given that

Speed of airplane,v=320 m/s

We have to find the smallest radius allowable for the vertical circle if the pilot's apparent weight is not exceed 7.0 times his true weight.

Let true weight =w

Apparent weight=7w

According to question

7w=w+\frac{mv^2}{r}

7w-w=\frac{mv^2}{r}

We know that

m=\frac{w}{g}

6w=\frac{wv^2}{rg}

6=\frac{v^2}{rg}

r=\frac{v^2}{6g}

Where g=9.8 m/s^2

r=\frac{(320)^2}{6\times 9.8}= 1741.5 m

r=\frac{1741.5}{1000}=1.74 km

Where 1 km=1000 m

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Explanation:

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A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu
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Answer:

please read the answer below

Explanation:

The angular momentum is given by

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b)

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L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

c)

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d)

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the same as e):

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g)

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h)

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the same as g):

L = 5kgm/s

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