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2 years ago

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A typical jet airliner has a cruise airspeed of 900 km/h , which is its speed relative to the air through which it is flying. If

the wind at the airliner’s cruise altitude is blowing at 100 km/h from west to east, what is the speed of the airliner relative to the ground if the airplane is flying from (a) west to east, and (b) east to west

1 answer:

Luba_88 [7]2 years ago

8 0

Explanation:

It is given that,

Speed of the jet airplane with respect to air, $v_{PA} = 900\ km/h$

If the wind at the airliner’s cruise altitude is blowing at 100 km/h from west to east, $v_{AG}= 100\ km/h$

(A) Let $v_{PG}$ is the speed of the airliner relative to the ground if the airplane is flying from west to east,

$v_{PG}=900-100=800\ km/h$

(B) Let $v'_{PG}$ is the speed of the airliner relative to the ground if the airplane is flying from east to west,

$v'_{PG}=900+100=1000\ km/h$

Hence, this is the required solution.

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Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

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Question 11 (1 point)

Ainat [17]

Answer:

gravitational potential energy.

Explanation:

Gravitational potential energy (GPE) can be defined as an energy possessed by an object or body due to its position above the earth surface.

Mathematically, gravitational potential energy is given by the formula;

$G.P.E = mgh$

Where,

G.P.E represents gravitational potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

This ultimately implies that, anytime there is height, the object must have gravitational potential energy.

Hence, an object possesses gravitational potential energy due to its height (position) and the earth's gravitational force.

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2 years ago

a person using a machine applies a force of 100 newton's over a distance of 10 Meters to raise a 500 n object 1.5 meters what is

Hoochie [10]

Answer:

Work done by the machine (W) = 500 × 1.5 = 750 J

Work supplied to the machine (W) = 100 × 10 = 1000 J

Here, work supplied to the machine is input work = 1000 J

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Why is accelration negative when velocity is positive.

Sindrei [870]

Answer:

An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

Explanation:

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2 years ago

Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?

noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

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