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lawyer [7]
1 year ago
8

Your flight takes you in the path of a large aircraft. in order to avoid the vortices you should fly

Physics
1 answer:
Reika [66]1 year ago
4 0

your flight takes you in the path of a large aircraft. in order to avoid the vortices, you should fly above the flight path of the large aircraft.

<h3>What is Vortex?</h3>
  • A vortex is an area in a fluid where the flow spins around an axis line, which may be straight or curved, according to the principles of fluid dynamics.
  • Smoke rings, whirlpools in a boat's wake, and the winds surrounding a tropical cyclone, tornado, or dust devil are examples of vortices that form in agitated fluids.
  • A significant element of turbulent flow is vortices. Vortices are defined by the distribution of velocity, vorticity (the curl of the flow velocity), and the idea of circulation.
  • The fluid flow velocity in most vortices is the highest close to the axis and drops in inverse proportion to the distance from the axis.

To learn more about vortex with the given link

brainly.com/question/14362952

#SPJ4

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A boy on the beach holds a spherical balloon filled with air. at 10:00am, the temperature on the beach is 20°c and the balloon h
Anon25 [30]
For idea gases, volume is directly proportional to temperature. That is, an increase in temperature leads to increase in volume and vice versa.

Therefore,
V1/T1 = V2/T2 => T2 = (V2*T1)/V1

Assuming that the balloon is spherical in shape,

V= 4/3*pi*R^3.... In the formula for calculating T2, 4/3*pi cancels out.

R1 = 30/2 15 cm; R2 = 30.5/2 = 15.25 cm; T1 = 20+273.15 =293.15 K

Therefore,

T2 = (R2^3*T1)/R1^3 = (15.25^3*293.15)/15^3 = 308.05 K = 34.9 °C
6 0
3 years ago
Determine the magnitude of the component of F directed along the axis of AB. Set F = 520 N .
Soloha48 [4]

Answer:

The component of F along AB is equal to Fcos45

F = 520N

Component along AB = 520cos45

= 367.7N

This is done by rotating the diamonds such that AB is now taken as the x-axis. Then the force F is resolved along AB.

Explanation:

7 0
3 years ago
Read 2 more answers
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
I NEED a LOT of help PLEASE!
Minchanka [31]
Infrared, visible light, then ultraviolet. Infrared is light that the human eye can not see and visible light is clearly light we can see then ultraviolet is has such a high frequency we can't see it either.
3 0
3 years ago
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What is the best inference for the speed of the car after 3 seconds
SVEN [57.7K]

Answer:

B. 17m/s

Explanation:

This question contains a graph that illustrates the relationship between the speed of a car over time. The graph shows that one can make an inference of the amount of time it takes for the car to cover a particular speed and vice versa.

In this case, after 3 seconds, the speed of the car will be 17 m/s. This inference was got by tracing the position of 3s in the x-axis to the value on the y-axis. Doing this, the best inference for the speed of the car after 3 seconds is 17m/s.

8 0
3 years ago
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