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erica [24]
3 years ago
10

The lifting force, F, exerted on an airplane wing varies jointly as the area, A, of the wing's surface and the square of the pla

ne's velocity, v. The lift of a wing with an area of 190190 square feet is 95009500 pounds when the plane is going 220220 miles per hour. Find the lifting force on the wing if the plane slows down to 190190 miles per hour. (Leave the variation constant in fraction form or round to at least 5 decimal places.
Physics
1 answer:
Mrrafil [7]3 years ago
3 0

Answer:

F'=708.53 N

Explanation:

We have,

The lifting force, F, exerted on an airplane wing varies jointly as the area, A, of the wing's surface and the square of the plane's velocity, v. It means tat,

F=kAv^2

k is constant

If, A = 190 Ft², v = 220 mph, F = 950 pounds

Let's find k first from above data. So,

k=\dfrac{F}{Av^2}\\\\k=\dfrac{950}{190\times 220^2}\\\\k=0.0001033

It is required to find the lifting force on the wing if the plane slows down to 190 miles per hour. Let F' is the new force. So,

F'=0.0001033\times 190\times (190)^2\\\\F'=708.53\ \text{pounds}

So, the lifting force is 708.53 pounds if the plane slows down to 190 miles per hour.

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Answer:

B

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The ball posses gravitational potential energy since it is held at a displacement to the ground ( zero point) and when released, the gravitational potential energy is converted to kinetic energy which leads to the fall of the ball until it is at zero displacement to the earth. The board likewise when bent to its maximum extent stored elastic potential energy as a result of the partial displacement of its constituent particle provided it is  not stretch beyond its elastic limit which can lead to deformation of the board and the elastic potential energy lost.

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4 years ago
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

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3 years ago
Can someone please help me with this physics question? I'm desperate!
Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

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u = The initial velocity of Jenny = 0

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20 = t²/2

t = √(20 × 2) = 2·√10

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d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

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v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

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