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Marina CMI [18]
3 years ago
7

Consider a jet on the deck of an aircraft carrier. In order to take off, a catapult is attached to the jet and it is flung off t

he deck. An F-18 has a gross weight of approximately 17,000 kg and must reach speeds of about 270 km/hr by the end of the 150 m deck. The catapults on aircraft carriers must be able to launch a variety of aircraft that have different masses and minimum air speeds for flight! How would you ensure that you provide enough energy to the aircraft to get it safely off the deck of the aircraft carrier?
Physics
1 answer:
Komok [63]3 years ago
4 0

Answer:

Energy = 47812500 [J]

Explanation:

This problem can be solved using the principle of energy and work. First we need to identify the input data that give us these are:

m = mass = 17000[kg]

for take-off the airplane needs a velocity of 270 [km/h] = 75[m/s]

distance = d = 150 [m]

Therefore we can calculathe the kinetic energy.

E_{k}=0.5*m*v^2\\E_{k}=0.5*17000*(75)^2\\E_{k}=47812500[J]

In this way we need a device that can develop the next Force, this force can be calculated using the working theorem.

W = F * d

F = 47812500 / 150

F = 318750 [N]

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Uranus is now completing its ____ orbit around the sun since its discovery.
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Se calienta para templar y endurecer, una llave española de acero de 200 gramos, elevando su temperatura hasta los 550°C y se in
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Answer:

La temperatura final es de aproximadamente 159,94°C

Explanation:

Los parámetros dados son;

La masa de la llave española de acero, m₁ = 200 gramos

La temperatura de la llave, T₁ = 550 ° C

La masa del recipiente de aluminio que contiene agua, m₂ = 250 gramos

La masa del agua en el recipiente de aluminio, m₃ = 220 gramos

La capacidad calorífica específica del hierro, C_{planchar}, c₁ = 0.499 ca/(g·°C)

La capacidad calorífica específica del aluminio, C_{Aluminio}, c₂ = 0.217 cal/(g·°C)

La capacidad calorífica específica del agua, C_{Agua}, c₃= 1 cal/(g·°C)

En equilibrio térmico, tenemos;

m₁·c₁·(T₁ - T) = m₂·c₂·(T -T₂) + m₃·c₃·(T - T₂)

Conectando los valores, da;

200 × 0.499 × (550 - T) = 250 × 0.217 × (T -18) + 220 × 1  × (T - 18)

Simplificando, usando una calculadora gráfica, obtenemos;

\dfrac{274450-499\cdot T}{5} = \dfrac{1097 \cdot T-19746}{4}

De también encontramos 'T' al convertirlo en el tema de la ecuación anterior aún usando una calculadora gráfica;

T = 1196530/7481 °C ≈ 159.94°C

La temperatura final,T ≈ 159.94°C.

8 0
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A long jumper leaves the ground at an angle of 20.0° above the
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</span>R = V²o <span>sin² 2(teta) / g  = 121 x 0.41 /9.8 = 5.06 m
</span>
2)<span>the maximum height reached
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3) for teta = 25°,  R = V²o sin² 2(teta) / g= 7.24m, H= V²o <span>sin² teta  / 2g=</span><span>1.1m
   same method for the others
</span><span>graph , look at the image, and do the others like the example
   </span>

5 0
4 years ago
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