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3 years ago

air passing over an airplanes wing travels ,and therefore exerts pressure.than air traveling beneath the wing.

1 answer:

5 0

Bernoulli's principle of laminar/lamellar air flow, I think. High flow speed = low pressure, low flow speed = high pressure I think. So, the wings/aerofoils are designed to induce a low pressure on the top side of the wing and a high pressure on the underside of the wing, thus producing an "aerodynamic upthrust" (a static upthrust comes from an object in water via Archimedes) and LIFT.

Two "particles" of air one going topside and the other underside meet again at the end of their motion across the wing. So, top side has to travel faster than bottom side. So top side has a lower "dynamic pressure" than underside.

And all that for 5 points ????????? (If I'm right, of course ... )

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30 points must be a legitimate answer and correct answer or I will report

kupik [55]

Answer:

it's C

Explanation:

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6 0

2 years ago

Read 2 more answers

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

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

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

3 0

3 years ago

If the magnitude of the resultant force is to be 500n, directed along the positive y axis, determine the magnitude of force f an

scoundrel [369]

The answer is 0=45.2 degrees

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3 years ago

I drop an egg from a certain distance and it takes the egg 3.74 seconds to reach the ground. How high up was the egg?

Ulleksa [173]

Answer:

Explanation:

<u>Free Fall Motion </u>

When a body is left to move in the air with no friction, the motion is ruled only by the force of gravity. The vertical distance a body travels in the air after a time t is .

$\displaystyle y=\frac{gt^2}{2}$