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likoan [24]
3 years ago
14

Suppose the pilot makes an emergency turn to avoid an approaching missile, subjecting himself to a centripetal acceleration of 1

0 gs, while flying at 670 m/s (supersonic). What is the radius, in km, of his turn? (This must be short-lived because fighter planes can only briefly endure such large accelerations without serious damage, and the pilot will soon black out at 10 gs.)
Physics
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

  R = 4580.61 m

Explanation:

given,

Speed of the flight, v = 670 m/s

centripetal acceleration, a_c = 10 g

                                         = 10 x 9.8 = 98 m/s²

Radius of his turn, R = ?

using centripetal acceleration

a_c = \dfrac{v^2}{R}

R = \dfrac{v^2}{a_c}

R = \dfrac{670^2}{98}

  R = 4580.61 m

Hence, The radius of the turn of the flight is equal to 4580.61 m.

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alexgriva [62]

Answer:

8000 kg/cm^3

Explanation:

The density formula is P = M/V

Density = P = ? expressed in kg/m^3

M = mass = 40,000 kg

V = volume = 5 m^3

P = 40,000/5

P = 8000 kg/m^3

6 0
3 years ago
What metal do they use inside the torch to conduct electricity
sammy [17]
Inside the torch, they usually use copper or brass to conduct electricity.
7 0
3 years ago
Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de
Vesnalui [34]

Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 \frac{m}{s}

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa.  Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido,  en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

  • A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.
  • v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

En este caso:

  • D1: 0.06 m
  • v1: ?
  • D2: 0.04 m
  • v2: 2.6 m/s

Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

8 0
3 years ago
An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.
Anna35 [415]

Answer

Given,

Energy absorbed, Q_H = 1.69\ kJ

Energy expels,Q_C =  1.25\ kJ

Temperature of cold reservoir, T = 27°C

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c) Power output

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   P = \dfrac{W}{t}

   P = \dfrac{0.44}{0.296}

   P = 1.486\ kW

8 0
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5 0
4 years ago
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