All categories

3 years ago

Which of the following would NOT have to be cleaned after each use?

Engineering

2 answers:

Artemon [7]3 years ago

8 0

Answer: O A.

Explanation:

Delicious77 [7]3 years ago

4 0

Answer:

OD.

nail brush

Explanation:

hope this will help you

You might be interested in

In the figure below, block A weighs 20 lb , while block B weighs 10 lb . Friction between the surfaces of the two blocks may be

scoray [572]

Answer:

As P is continually increased, the block will now slip, with the friction force acting on the block being: f = muK*N, where muK is the coefficient of kinetic friction, with f remaining constant thereafter as P is increased.

8 0

3 years ago

9. Cigarette smoking is dangerous to one's health. Which of the following is the

Alla [95]

C.nicotine
The worse compound

5 0

3 years ago

Read 2 more answers

Steam enters an adiabatic turbine at 800 psia and9008F and leaves at a pressure of 40 psia. Determine themaximum amount of work

Naily [24]

Answer:

$w_{out}=319.1\frac{BTU}{lbm}$

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

$h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}$

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

$s_2=s_1$

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

$x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965$

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

$h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}$

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

$h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}$

Best regards.

3 0

3 years ago

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 75 MPa (68.25 ksi). If the plate is

wariber [46]

Answer:

Explanation:

From the given information:

Strain fracture toughness $K_k$ = 75 MPa $\sqrt{m}$

Tensile stress $\sigma$ = 361 MPa

Value of Y = 1.03

Thus, the minimum length of the critical interior surface crack which will result to fracture can be determined by using the formula:

$a_c = \dfrac{1}{\pi} ( \dfrac{k_k}{\sigma Y})^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ \dfrac{75 \times \sqrt{10^3}}{361 \times 1.03 } \Big]^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ 6.378474693\Big]^2 \\ \\ \mathbf{ a_c = 12.95 \ mm}$