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Alex
3 years ago
10

At cruise conditions, air flows into a jet engine at a steady rate of 60 lbm/s. Fuel enters the engine at a steady rate of 0.59

lbm/s. The average velocity of the exhaust gases is 1485 ft/s relative to the engine. If the engine exhaust effective cross-sectional area is 4.0 ft2, estimate the density of the exhaust gases in lbm/ft3.
Physics
1 answer:
Amanda [17]3 years ago
8 0

Answer:

ρ=0.0102lbm/ft^3

Explanation:

To solve this problem we must take into account the equation of continuity, this indicates that the sum of the mass flows that enter a system is equal to the sum of all those that leave.

Therefore, to find the mass flow of exhaust gases we must add the mass flows of air and fuel.

m=0.59+60=60.59lbm/s( mass flow of exhaust gases)

The equation that defines the mass flow (amount of mass that passes through a pipe per unit of time) is as follows

m=ρVA

Where

ρ=density

V=velocity

m=mass flow

A=cross-sectional area

solving for density

ρ=m/VA

ρ=60.59/{(1485)(4)}

ρ=0.0102lbm/ft^3

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Which of the following sets of characteristics describe what we know about the outer planets? (2 points)
masha68 [24]

Answer: D

Explanation: Gaseous composition, larger size and many moons

6 0
3 years ago
If the satellite has a mass of 3900 kg , a radius of 4.3 m , and the rockets each add a mass of 210 kg , what is the required st
ehidna [41]

Answer:

The required steady force of each rocket is 28.79 N

Explanation:

mass of the satellite, M=3900 kg

radius, r=4.3 m

mass of rocket, m=210 kg

time, t=5.4 min

Moment of Inertia:

I = 1/2 (Mr^2) + 4mr^2

I = 1/2 ( 3900* (4.3)^2) + 4 (210)*(4.3)^2

I = 51587.1 kg m^2

the angular acceleration is:

a= w/t

here w= 2*π*30

so,

a= 2*π*30 / 5.4* 3600

a=0.0096 rad/ s^2

the Torque becomes:

T=I*a = 4r*F

( 51587.1 )*(0.0096) = 4*4.3* F

F= 28.79 N

the required steady force of each rocket is 28.79 N

learn more about steady force here:

<u>brainly.com/question/13841147</u>

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7 0
2 years ago
A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

5 0
3 years ago
Which list places the layers of the sun in the correct order from outermost to innermost?
Valentin [98]
Answer - corona, chromosphere, photosphere
4 0
3 years ago
A student throws a baseball at a large gong 52 m away and hears the sound of the gong 1.73333 s later. The speed of sound in air
Arte-miy333 [17]

Speed = 52/1.57576= 33 m/s

Explanation:

Distance = speed * time

Given , the distance traveled by the baseball = 52 m

Speed of sound in air = 330 m/s

Total time  = 1.73333 s .

Total time for the student to hear the sound of the gong is the sum of the time take for the baseball to reach the gong and the time taken by the sound to travel back.

Distance traveled by the sound is 52 m and the speed is 330 m/s

So time taken by the sound to travel back = distance traveled / the speed

=> time = 52/330 s = 0.15757 s

Time taken by the base ball to reach the gong is the total time - the time taken by the sound

=> time taken by base ball = 1.73333 - 0.15757 = 1.57576s

Speed of the base ball to reach the gong = distance / time

Speed = 52/1.57576= 33 m/s

3 0
3 years ago
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