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3 years ago

NASA SPACE SHUTTLE QUESTION:

Engineering

1 answer:

vfiekz [6]3 years ago

8 0

Explanation:

The Challenger accident made them aware of the risks. They had been a little naive, since it was never believed that such a thing could happen. This, together with Space Shuttle Columbia disaster revealed the risks of the space shuttle program, which also had very high maintenance costs.

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How many hours should it take an articulated wheel loader equipped with a 4-yd^3 bucket to load 3000 yd^3 of gravel (average mat

densk [106]

Answer:

17 hours 15 minutes

Explanation:

See attached picture.

4 0

3 years ago

Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1

babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0

3 years ago

True or false a critique of hazwoper incidents that have occurred in the past year should not be included in hazwoper 8 hour ref

Jobisdone [24]

Answer:

False

Explanation:

No matter if something happened in the past year or so, it still should be included for safety reasons so it wont happen again

7 0

3 years ago

Bài 3: Cho cơ cấu culít (hình 3.5) với các kích thước động lAB = 0,5lAC = 0,1m. Khâu 3 chịu tác dụng của mô men M3 = 500 N. Cơ c

lesya [120]

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3 0

2 years ago

As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial tempera

saveliy_v [14]

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod $\overline T = 548 \ K$

$\rho = 7900 \ kg/m^3$

$K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K$

$\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657$

Here, we can't apply the lumped capacitance method, since Bi > 0.1

$\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\$

$0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81$

$t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins$

However, on a single rod, the energy extracted is:

$\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J$

Hence, for centerline temperature at 50 °C;

The surface temperature is:

$T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}$

5 0

3 years ago