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Aleks04 [339]
3 years ago
14

When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing

is 273 mph relative to the airplane. (a) What suction pressure is developed on the wing at that point? (b) What is the pressure at the leading edge (a stagnation point) of the wing?
Physics
1 answer:
Sedaia [141]3 years ago
5 0

Answer:

P1 = 0 gage

P2 = 87.9 lb/ft³

Explanation:

Given data

Airplane flying = 200 mph = 293.33 ft/s

altitude height = 5000-ft

air velocity relative to the airplane = 273 mph = 400.4 ft/s

Solution

we know density at height 5000-ft is 2.04 × 10^{-3} slug/ft³

so here P1 + \frac{\rho v1^2}{2}  = P2 + \frac{\rho v2^2}{2}

and here

P1 = 0 gage

because P1 = atmospheric pressure

and so here put here value and we get

P1 + \frac{\rho v1^2}{2}  = P2 + \frac{\rho v2^2}{2}

0 + \frac{2.048 \times 10^{-3} \times 293.33^2}{2}  = P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}  

solve it we get

P2 = 87.9 lb/ft³

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A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward
AleksandrR [38]

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

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5 0
3 years ago
A 500g object falls off a cliff and losers 100 J from its gravitational potential energy store. if the gravitational field stren
ludmilkaskok [199]

Answer : Height, h = 20.4 m

Explanation :

It is given that,

Mass of an object, m = 500 g = 0.5 kg

Gravitational potential energy, PE = 100 J

The Gravitational potential energy is the energy which is possessed due to the height and gravity of an object. It is given as :

PE = m g h

where,

h is the height of the cliff.

100\ J=0.5\ kg\times 9.8\ m/s^2\times h

h = 20.40 m

So, the height of the cliff is 20.4 m.

7 0
3 years ago
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