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1 year ago

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A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising th

e helicopter. An 810-kg helicopter rises from rest to a speed of 7.0 m/s in a time of 3.5 s. During this time it climbs to a height of 8.2 m. What is the average power generated by the lifting force?
a) M = 810 kg
b) V0 = 0 m/s
c) Vf = 7 m/s
d) t = 3.5 s
e) y = 8.2 m

1 answer:

kobusy [5.1K]1 year ago

7 0

Answer:

24,267.6 watts

Explanation:

from the question we are given the following:

mass (m) = 810 kg

final velocity (v) = 7 m/s

initial velocity (u) = 0 m/s

time (t) = 3.5 s

final height (h₁) = 8.2 m

initial height (h₀) = 0 m

acceleration due to gravity (g) = 9.8 m/s^{2}

find the power

power = \frac{work done}[time}

and

work done = change in kinetic energy (K.E) + change in potential energy (P.E)

work done = (0.5 mv^{2} - 0.5 mu^{2} ) + ( mgh₁ - mgh₀)

since u and h₀ are zero the work done now becomes

work done = (0.5 mv^{2}) + ( mgh₁ )

work done = (0.5 x 810 x 7^{2}) + ( 810 x 9.8 x 8.2)

work done = 84, 936.6 joules

recall that power = \frac{work done}[time}

power = \frac{84,936.6}[3.5}

power = 24,267.6 watts

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C: 5000 N

D: 84,8 J

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A.

In the first question, we have to caculate the constant of the spring with this equation:

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Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

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The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

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The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

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