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3 years ago

What effect does air have on the acceleration of aircraft during flight?

1 answer:

5 0

The effect would be the altitude of the air, the higher you go up the closer you are to space we’re there’s no oxygen and everything moves slow so when your trying to fly across the world it could feel like your moving slower

You might be interested in

Which of the following is NOT true about hydraulic systems?

Dmitry [639]

Answer: The answer is D

D.In hydraulic systems, the operating temperatures must be kept between 170�F and 180°F

Explanation:

The operating temperature for hydraulic systems is 140°F and below. Anything above this temperature is too high and will reduce the useful life of hydraulic fluid.

Most often problems associated with hydraulic systems are caused by fluid contaminated with particulate matter.

7 0

3 years ago

Type the correct answer in the box. Spell all words correctly.

Rudik [331]

Answer:ii dant overstand

Explanation:

5 0

3 years ago

When they say in the United States that a car’s tire is filled “to 32 lb,” they mean that <br>its internal pressure is 32 lbf/i

arsen [322]

Answer:

0.71 lbf

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.

P = 32 lbf/in² + 14.7 lbf/in²

P = 46.7 lbf/in²

Absolute temperature is in Kelvin or Rankine:

T = 75 + 459.67 R

T = 534.67 R

Given V = 3.0 ft³, and R = 10.731 ft³ psi / R / lb-mol:

PV = nRT

(46.7 lbf/in²) (3.0 ft³) = n (10.731 ft³ psi / R / lb-mol) (534.67 R)

n = 0.02442 lb-mol

The molar mass of air is 29 lbm/lb-mol, so the mass is:

m = (0.02442 lb-mol) (29 lbm/lb-mol)

m = 0.708 lbm

The weight of 1 lbm is lbf.

W = 0.708 lbf

Rounded to two significant figures, the weight of the air is 0.71 lbf.

3 0

3 years ago

A thin rim with a mean diameter of 1.2 m cross-section of 15 mm x 200 mm is subjected to an internal pressure of 10 MPa and rota

Soloha48 [4]

Answer:

The net centrifugal force over the rim is 30000N, the radial stress is 397887 Pa and the total change in diameter is 4.98 mm.

Explanation:

Lets first calculate the force in the rim due to the centrifugal force. For doing this we may assume that the centrifugal force is constant along with thick because of the thin thick.

Fc = m.ω^2/R

Where m is the mass, w the angular speed and R the mean radius. The mass is computing by the rim density and its volume:

m=p.V

m=p*(A*R)

Where A is the cross-sectional area in meters:

m=((0.015m*0.200m)*0.6m)*(7800 kg/m^3)=28.08 kg

The angular speed in rad/s is:

ω = 800r/m . 1m/60s = 133.33 r/s

Thus the centrifugal force is:

Fc = (28.08 kg)*(133.33 rad/s)^2*(0.6m) = 299505N = 30000N

Note that the calculating value is the net contribution to the whole rim but the centrifugal force is distributed along the rim's external area:

fc = Fc / (2π .R .b)

Where b is rim's with equal to 200mm :

fc = 300000 N / (2π*0.6m*0.2m) = 397887 N/m^2

The centrifugal force can be taken as internal pressure:

Pfc = 397887 N/m^2 = 3978787 Pa

As both pressures act expanding the rim it can be summed:

Pt=Pi+Pfc

Pt = 10MPa+397887Pa= 10000000Pa+397887Pa= 10397887Pa

Then for a thinner thick the stress is calculated by:

Pt*d =2σr*t

Take into account that the stress σr is over the radial direction. Then solving for o and by replacing the total pressure:

σr = Pt.d/(2*t)

σr = 10397887 Pa / (2*0.015m*0.2m) = 415915480Pa = 415MPa

We know that the radial specific deformation ε is:

σr = E / εr

εr = σr / E

For a young modulus of 200GPa:

εr = 415MPa / 200GPa

εr = 415MPa / 200000MPa=0.002075

By definition the specific deformation is written in terms of the total change in the radius:

εr = Δr / R

Δr = R / εr =0.002075 * 1.2 m = 0.00249m

As we need the change in diameter:

Δd = 2Δr =0.00498m= 4.98mm [/tex]

5 0

3 years ago

Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.3

Paladinen [302]

Answer:

A) n = 0.6143, c ≈ 640m/min

B) n = 0.6143 , c = 637.53m/min

Explanation:

using the given data

A) A plot of flank wear as a function of time and also A plot for tool when

Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min also for cutting edge speed of 155m/min , time for cutting edge is 10 min

is attached below

calculate for the constant N from the second plot

note : the slope will be negative because cutting speed decreases as time of cutting increase

V1 = 100m/min , V2 = 155m/min, T1 = 20.4 min, T2 = 10 min

= - N = $\frac{In(V2) - In(V1)}{In(T2)-ln(T1)}$

therefore - N = $\frac{5.043 - 4.605}{2.302 -3.015}$

= - 0.6143

THEREFORE ( N ) = 0.6143

Determine for the constant C from the second plot as well

note : C is the intercept on the cutting speed axis in 1 min tool life

connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point

hence C ≈ 640m/min

B) Calculate the values of N and C in the Taylor equation solving simultaneous equations

using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation

Taylor tool life equation = vT = C ------------- equation 1

cutting speed = v = 100m/min and 155m/min

tool life = T = 20.4 min and 10 min

also constant n and c are obtained from the previous plot

back to taylor tool life equation = 100 * 20.4 = C

therefore C = (100)(20.4)^n ---------------- equation 2

also using the second values of v and T

taylor tool life equation = 155 * 10 = C

therefore C = ( 155 )(10)^n ----------------- equation 3

Equate equation 2 and equation 3 and solve simultaneously

(100)(20.4)^n = (155)(10)^n

To find N

take natural log of both sides of the equation

= In ((100)(20.4)^n) = In((155)(10)^n)

= In (100) + nIn(20.4) = In(155) + nIn(10)^n

= n(3.0155) - n (2.3026) = 5.043 - 4.605

= 0.7129 n = 0.438

therefore n = 0.6143

To find C

substitute 0.6143 for n in equation 2

C = (100)(20.4) ^ 0.6143

C = 637.53 m/min

Attached are the two plots for solution A

7 0

3 years ago