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3 years ago

What does an aeronautical engineer design

Engineering

1 answer:

Colt1911 [192]3 years ago

3 0

Answer:

they work with aircraft, designing aircrafts.

Explanation:

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An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8

marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

P= 800 x 108.3

P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

$a=18.66\ m/s^2$

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0

3 years ago

A debugging process where you, the programmer, pretend you are a computer and step through each statement while recording the va

Ipatiy [6.2K]

Answer:

hand tracing

Explanation:

as a programmer when we pretend computer in the debugging process by the step of each statement in recording

then there value of variable is hand tracing because as The hand tracking feature is the use of hands as an input method

so while recording value of each variable each step is hand tracing

5 0

3 years ago

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and

k0ka [10]

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

$z1=$ $\frac{\left(150-137\right)}{27.7}$ $=0.4693$

$z2=\frac{\left(201-137\right)}{27.7}=2.3105$

$P(0.4693$

B.)

$z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289$

$\\P(2.9309$

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

3 0

3 years ago

Read 2 more answers

Write 83,120 in expanded form using powers of 10.

maksim [4K]

Answer:

8*10000+3*1000+1*00+2*10+2

Explanation:

8 0

2 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

max2010maxim [7]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

7 0

3 years ago