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3 years ago

6

An airplane is flying in a horizontal circle at a speed of 480 km/h (). If its wings are tilted at angle =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.

1 answer:

german3 years ago

6 0

Answer:

$R = 2162 m$

Explanation:

When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components

$F_y = mg$

$F_x = \frac{mv^2}{R}$

now we know that

$F_y = F cos40$

$F_x = F sin40$

also we know that

$v = 480 km/h$

$v = 133.3 m/s$

now plug in all data in above equations

$tan 40 = \frac{v^2}{Rg}$

$R = \frac{v^2}{g tan40}$

$R = \frac{133.3^2}{9.8 tan40}$

$R = 2162 m$

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7 0

3 years ago

Read 2 more answers

Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.

DaniilM [7]

Answer: $2(10)^{-9} mg$

Explanation:

We know the total diameter of the cell (assumed spherical) is:

$d=1.9\mu m=1.9(10)^{-6} m$

Then its total radius $r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m$

On the other hand, we know the thickness of the cell wall is $r_{t}=60 nm= 60(10)^{-9} m$ and its density is the same as water ( $\rho=997 kg/m^{3}$ ).

Since density is the relation between the mass $m$ and the volume $V$ :

$\rho=\frac{m}{V}$

The mass is: $m=\rho V$ (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

$V=\frac{4}{3}\pi R^{3}$ (2)

Where $R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m$

Then:

$V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3}$ (3)

$V=2.952(10)^{-18} m^{3}$ (4)

Substituting (4) in (1):

$m=(997 kg/m^{3})(2.952(10)^{-18} m^{3})$ (5)

$m=2.94(10)^{-15} kg$ (6)

Knowing $1 kg=1000 g$ and $1 mg=0.001 g$ :

$m=2.94(10)^{-15} kg=2(10)^{-9} mg$

7 0

4 years ago

An airplane in the process of taking off travels with a speed of 80 m/s at an angle of 15° above the horizontal. What is the gro

9966 [12]

Answer:

Option C

Explanation:

given,

velocity of airplane = 80 m/s

angle with the horizontal = 15°

speed of the ground= ?

when the plane is taking off the horizontal component of the velocity is v cosθ

so,

ground speed of the airplane is = $v\times cos\theta$

= $80 \times cos 15^0$

v = 77.27 m/s

horizontal velocity of the air plane comes out to be 77.27 m/s ≅ 77 m/s

so, the correct option is Option C

5 0

4 years ago