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3 years ago

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A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee

d of 40.0 m/s, the cable makes an angle of 38.0° with respect to the vertical. Determine the force exerted by air resistance on the bucket.

1 answer:

rodikova [14]3 years ago

5 0

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water = 700-kg

length of cable = 20 m

Speed = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

$T sin 38^0= \dfrac{mv^2}{l} + F$ ..........(2)

equation (1)/(2)

$tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}$

$0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}$

F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

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$\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}$ .

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In other words, the work input of the ideal lever is equal to the work output.

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