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lisov135 [29]
3 years ago
8

An airplane is flying in a horizontal circle at a speed of 480 km/h. If its wings are tilted 40° to the horizontal, what is the

radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.
Physics
1 answer:
guajiro [1.7K]3 years ago
3 0

Answer:

r = 2161.9 m

Explanation:

Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.

Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.

L is perpendicular to wing at angle θ with respect to horizontal

Thus,

Vertical component of lift is:

L cosθ = W = mg

Thus, m = L cosθ / g - - - - (eq1)

Horizontal component of lift is:

L sinθ = centripetal force = mv² / r - - - - (eq2)

Combining equations 1 and 2,we have;

L sinθ = (L cosθ / g)(v² / r)

L cancels out on both sides to give;

tanθ = v²/ rg

r = v² / (g tanθ)

We are given;

velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s

r = 133.33²/[(9.8) tan(40)] = 2161.9 m

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Natasha_Volkova [10]

Answer:

for what?

Explanation:

d=S x T

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d=vt+1/2at2

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8 0
3 years ago
Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
Se necesita subir una carga de 500 kg (4900 N) a una altura de 1.5 m deslizándola sobre una rampa inclinada. ¿Qué longitud debe
marusya05 [52]

Answer:

4.22 m

Explanation:

Una rampa es una máquina que se utiliza para levantar un objeto con una fuerza menor a la que realmente necesitarías. Cuanto mayor sea la longitud de la rampa, menor será la magnitud de la fuerza necesaria para levantar el objeto.

Dado que:

altura de la rampa = 1.5 m, carga = 4900 N, fuerza aplicada = 1633.33 N.

La fórmula de la rampa se da como:

fuerza aplicada * longitud de la rampa = peso de la carga * altura de la rampa

1633.33 * longitud de la rampa = 4900 * 1.5

longitud de la rampa = 4900 * 1.5 / 1633.33

longitud de la rampa = 4.22 m

6 0
3 years ago
What is the potential energy of a 2500 g object suspended 5 m above the earth's surface?
alexandr402 [8]

Answer:

Explanation:

Potential energy, which is the energy a body assumes at a position, can be calculated using the formula:

P.E = m × g × h

Where;

m = mass (kg)

g = acceleration due to gravity (10m/s²)

h = height (m)

7 0
3 years ago
Water flows over a section of Niagara Falls at the rate of 1.4 × 106 kg/s and falls 49.8 m. How much power is generated by the f
Ad libitum [116K]

Answer:

Power= 6.84×10⁸ W

Explanation:

Given Data

Niagara falls at rate of=1.4×10⁶ kg/s

falls=49.8 m

To find

Power Generated

Solution

Regarding this problem

GPE (gravitational potential energy) declines each second is given from that you will  find much the kinetic energy of the falling water is increasing each second.

So power can be found by follow

Power= dE/dt = d/dt (mgh)

Power= gh dm/dt

Power= 1.4×10⁶ kg/s × 9.81 m/s² × 49.8 m

Power= 6.84×10⁸ W  

7 0
3 years ago
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