I need help with 5 please and thank you

the idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse

Finger [1]

Forward thrust has positive values and reverse thrust has negative values.

Thrust is a sudden push or pull in a certain direction.

a)

Flight speed u = 150 km/h

1 km/h = $\frac{1}{3.6}$ km/s

therefore, 150 km/h = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

$T = m_{exhaust} * U_{exhaust} - U_{flight}$

T = 50 * (150 - 41.67)

T = 5416.67 N

Therefore, the value of forward thrust is 5416.67 N.

b)

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component,

thus thrust equation is:

$T = m_{exhaust} * U_{exhaust} - U_{flight}$

T = 50 * (0 - 41.67)

T = -2083.5 N

Therefore, the thrust force T is -2083.5 N in the reverse direction.

c)

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero because $U_{exhaust} = U_{flight} = 0\\$

T = 0

Therefore, there is no difference in two velocities in x direction.

The given question is incomplete, the complete question is,

"The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.

a. What is the forward thrust of this engine?

b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?

c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?"

To know more about thrust,

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3 0

1 year ago

In a circuit of 2 lamps in parallel if the current in one lamp is 2a the current in the other lamp is?

ioda

In a circuit having 2 lamps are connected in parallel to a battery

then the two lamps will be having the same potential as the battery

i.e

$V_{1} = V_{2} = V_{battery}$

As per Ohm's law,

$I_{1} = \frac{V_{1}}{R_{1}}$ and $I_{2} = \frac{V}{R_{2} }$

In other words, each lamp's current is inversely related to its individual resistance. We only know the current in one of the bulbs in this specific instance. We would therefore need further information in order to calculate the current in the other light. Therefore, there isn't enough data to make a statement.

Under the assumption that all physical parameters, including temperature, remain constant, Ohm's law asserts that "the voltage across a conductor is directly proportional to the current flowing through it".

Learn more about Ohm's law here

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4 0

1 year ago

What is the resistance of a bulb of 4ow connected in a line of 220v? 2

inn [45]

Answer:

1210 ohm

Explanation:

Given :

P=40 W

V=220 V

Now,

$P=\frac{V^{2} }{R} \\40=\frac{(220)^{2} }{R} \\40R=48400\\R=\frac{48400}{40} \\R=1210 ohm$

Therefore, resistance of bulb will be 1210 ohm

8 0

3 years ago

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24}kg$ is the mass of the Earth

$r$ is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period $T_{X}$ is:

$T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}$ (2)

Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

For Satellite X:

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

For Satellite Y:

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0

3 years ago

Read 2 more answers

Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0

lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

3 0

3 years ago