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nikklg [1K]
4 years ago
14

An airplane is cruising at an altitude of several kilometers. The pressure outside the craft is 0.258 atm within the passenger c

ompartment the pressure is 1.00 atm and the temperature is 20∘C. The density of air is 1.20 kg/m3 at 20∘C and 1 atm of pressure. A small leak occurs in one of the window seals in the passenger compartment. Model the air as an ideal fluid to find the speed of the stream of air flowing through the leak.

Physics
1 answer:
garik1379 [7]4 years ago
8 0

Answer:

<em>The speed of the stream of air flowing through the leak is 340.754 m/sec</em>

Explanation:

Carefully applying the Bernoulli's equation the speed of the leak can be obtained. The attached images show step by step explanation of the question, while applying the Bernoulli's equation;

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The Hubble Space Telescope is in a "Low-Earth Orbit" with an orbital period of 95min. Calculate the altitude of the HST's orbit
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A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What i
Arte-miy333 [17]

(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

K=\frac{1}{2}mv^2

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

v^2 = u^2 + 2gd

where

u = 21 m/s is the initial speed of the rock as it enters the hole

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Substituting,

v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s

(b) 4.72 s

The vertical position of the rock at time t is given by

y(t) = v_y t - \frac{1}{2}gt^2

where

v_y = 21 m/s is the initial vertical velocity

Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:

-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0

which has two solutions:

t = -0.43 s --> negative, so we discard it

t = 4.72 s --> this is our solution

7 0
4 years ago
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