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Sindrei [870]
3 years ago
9

Answer this question, please​

Physics
1 answer:
pashok25 [27]3 years ago
3 0

Answer:

the correct answer is the 60

+ 20 + 60 \gamma  \beta

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Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density =  0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

  =  -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level )  = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = \sqrt{CRT}  =  \sqrt{1.4*287*200.5 } = 283.8 m/s

hence it is a subsonic aircraft

4 0
3 years ago
Which statement best describes frustration?
wlad13 [49]
C. when you can't achieve your goal due to events beyond your control
5 0
3 years ago
A train travels 8.81 m/s in a -51.0° direction.
Amiraneli [1.4K]

The displacement of the train after 2.23 seconds is 25.4 m.

<h3>Resultant velocity of the train</h3>

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

  • θ is the angle between the velocity = (90 - 51) + 37 = 76⁰

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

R = 11.39 m/s

<h3>Displacement of the train</h3>

Δx = vt

Δx = 11.39 m/s x 2.23 s

Δx = 25.4 m

Thus, the displacement of the train after 2.23 seconds is 25.4 m.

Learn more about displacement here: brainly.com/question/2109763

#SPJ1

8 0
1 year ago
What causes the coriolis effect
Marta_Voda [28]
C is the correct answer
7 0
3 years ago
Two 0.2304 cm x 0.2304 cm square aluminum electrodes, spaced 0.5974 mm apart are connected to a 61 V battery. What is the charge
Arisa [49]

Answer:

The charge on each plate is 0.0048 nC

Explanation:

for the distance between the plates d and given the area of plates, A, and ε = 8.85×10^-12 C^2/N.m^2, the capacitance of the plates is given by:

C = (A×ε)/d

   =[(0.2304×10^-2)(0.2304×10^-2)×(8.85×10^-12))/(0.5974×10^-3)

   = 7.86×10^-14 F

then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:

q = C×V

   = (7.86×10^-14)×(61)

   = 4.80×10^-14 C

   ≈ 0.0048 nC

Therefore, the charge on each plate is 0.0048 nC.

3 0
3 years ago
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