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3 years ago

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A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b

e airborne after a take off run of 208 m?

1 answer:

netineya [11]3 years ago

6 0

Answer:

The acceleration is $a =51945 \ km/h^2$

Explanation:

From the question we are told that

The lift up speed is $v = 147 \ km/h$

The distance covered for the take off run is $s = 208 m = 0.208 \ km$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest

So

$147^2 = 0^2 + 2* a* 0.208$

=> $a =51945 \ km/h^2$

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Explanation:

For this case we have the following data given:

$W= 71.2 N$ represent the weigth for the object

$R=3.8 m$ the length of the rope or the radius

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Part a

For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

$a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2$

Part b

For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

$T= mg +ma$

In order to find the mass we know that $W=mg$ and solving for m we got:

$m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg$

And replacing into the tension we got:

$T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N$

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Answer:

Explanation:

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R:radius of the disk

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G: Cavendish's constant

by solving the integral you obtain:

$F_x=2\frac{GmM}{R^2}[1-cos\theta]$

$F_x=2GmM[1-\frac{a}{\sqrt{R^2+a^2}}]$ (1)

To find the gravitational energy you use:

$U=-\int F_x dx=-\int[2GmM(1-\frac{x}{\sqrt{R^2+x^2}})]dx\\\\U=-2GmM[x+\sqrt{R^2+x^2}]\\\\U=-2GmM[a+\sqrt{R^2+a^2}]$

you replace the values of the parameters in the point A:

$U=-2(6.67*10^{-11})(250)(5.98*10^{24})[(7.83*10^6)+(\sqrt{(7.83*10^6)^2+(6.25*10^7)^2})]\\\\U=-1.41*10^{25}J$

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However, the net work is also the difference between the gravitaional potential energies calculated in the previous steps.

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3 years ago