All categories

2 years ago

What are the career pathways for Aerospace engineering?

1 answer:

6 0

1. mechanical engineer, aircraft/space craft designer. Data processing manager. Military aerospace engineer. Inspector and compliance officer. Drafter. Aerospace technician. Mission or payload specialist.

2. ability to develop and conduct appropriate experimentation analyze and interpret data and ability to acquire in a play new knowledge is needed using appropriate learning strategies

3. Graduates of the co-op program are also up and offered a higher starting salary the co-op program for aerospace engineering students provides one full year of work experience divided into three segments of fall semester spring semester and a summer session

4. 4 years. no according to bachelors portal to join aerospace as an engineer or scientist you will have to study the subject for 4 to 7 years after high school

You might be interested in

A train which is traveling at 70 mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its

Ugo [173]

Answer:

a) 0 mi/s^2

b) 52 mi/s

Explanation:

Assuming the crossing is 1/2 mile past point A and that point B is near point A (it isn't clear in the problem)

The train was running at 70 mi/h at point A and with constant deceleration reachesn the crossing 1/2 mile away with a speed of 52 mi/h

The equation for position under constant acceleration is:

X(t) = X0 + V0 * t + 1/2 * a * t^2

I set my reference system so that the train passes point A at t=0 and point A is X = 0, so X0 = 0.

Also the equation for speed under constant acceleration is:

V(t) = V0 + a * t

Replacing

52 = 70 + a * t

Rearranging

a * t = 52 - 70

a = -18/t

I can then calculate the time it will take it to reach the crossing

1/2 * a * t^2 + V0 * t - X(t) = 0

Replacing

1/2 (-18/t) * t^ + 70 * t - 1/2 = 0

-9 * t + 70 * t = 1/2

61 * t = 1/2

t = (1/2)/61 = 0.0082 h = 29.5 s

And the acceleration is:

a = -18/0.0082 = -2195 mi/(h^2)

To beath the train the car must reach the crossing in 29.5 - 4.3 = 25.2 s

X(t) = X0 + V0 * t + 1/2 * a * t^2

52 mi/h = 0.0144 mi/s

1/2 = 0 + 0.0144 * 25.2 + 1/2 * a * 25.2^2

1/2 = 0.363 + 317.5 * a

317.5 * a = 0.5 - 0.363

a = 0.137/317.5 = 0.00043 mi/s^2 (its almost zero)

The car should remain at about constant speed.

It will be running at the same speed.

4 0

3 years ago

Which process is a from of mechanical weathering

alekssr [168]

Ice. The formation of ice in the myriad of tiny cracks and joints in a rock's surface slowly pries it apart over thousands of years. Frost wedging results when the formation of ice widens and deepens the cracks, breaking off pieces and slabs. Frost wedging is most effective in those climates that have many cycles of freezing and thawing. Frost heaving is the process by which rocks are lifted vertically from soil by the formation of ice. Water freezes first under rock fragments and boulders in the soil; the repeated freezing and thawing of ice gradually pushes the rocks to the surface.

7 0

3 years ago

Calculate the number of vacancies per cubic meter for some metal, M, at 783°C. The energy for vacancy formation is 0.95 eV/atom,

djyliett [7]

Answer:

Following are the solution to this question:

Explanation:

The number of vacancies by the cubic meter is determined.

$N_V =N exp(\frac{Q_v}{kT})$

$= \frac{N_A \rho}{A} exp (\frac{Q_v}{kT})$

$= \frac{6.022 \times 10^{23} \times 6.10}{43.41} \exp(\frac{-0.95}{8.62\times 10^{-5} \times (783+273)})\\\\= \frac{36.7342 \times 10^{23}}{43.41} \exp(\frac{-0.95}{0.0313626})\\\\= 0.846215158 \times 10^{23} \exp(-30.290856)\\\\$

$=1.57 \times 10^{25} \ cm^{-3}$

7 0

3 years ago

The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

DENIUS [597]

Answer:

See below

Explanation:

Check One-Sample T-Interval Conditions

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

One-Sample T-Interval Information

Formula --> $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})$
Sample Mean --> $\bar{x}=390.25$
Critical Value --> $t^*=2.0096$ (given $df=n-1=50-1=49$ degrees of freedom at a 95% confidence level)
Sample Size --> $n=50$
Sample Standard Deviation --> $S_x=30.5$

Problem 1

The critical t-value, as mentioned previously, would be $t^*=2.0096$ , making the 95% confidence interval equal to $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}$

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0

2 years ago

One cylinder in the diesel engine of a truck has an initial volume of 650 cm3 . Air is admitted to the cylinder at 35 ∘C and a p

kupik [55]

Answer:

1) the final temperature is T2 = 876.76°C

2) the final volume is V2 = 24.14 cm³

Explanation:

We can model the gas behaviour as an ideal gas, then

P*V=n*R*T

since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:

P*V^k = constant = C, k= adiabatic coefficient for air = 1.4

then the work will be

W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)

W = (P1*V1/T1)*(T2-T1)/(1-k)

T2 = (1-k)W* T1/(P1*V1) +T1

replacing values (W=-450 J since it is the work done by the gas to the piston)

T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C

the final volume is

TV^(k-1)= constant

therefore

T2/T1= (V2/V1)^(1-k)

V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³

3 0

3 years ago