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bija089 [108]
3 years ago
14

What is the resolution limit for a projection type of photolithographic system if the incident wavelength is 365 nm (the i-line

of Hg), N.A.=0.63, and k1=0.6. what is the Depth of field needed for the best resolution?
Engineering
1 answer:
Vadim26 [7]3 years ago
7 0

Answer:

Depth of field  = 347.619 nm

Explanation:

wavelenght =  365nm

N.A =0.63

k1= 0.6

so we have that the resolution limit is:

R=k1*A

R=0.6*365

R=219 nm

and the Depth of field needed for the best resolution is:

DoF = Resolution / N.A.

DoF= R/N.A

DoF= 219/0.63

DoF= 347.619 nm

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A circular hoop sits in a stream of water, oriented perpendicular to the current. If the area of the hoop is doubled, the flux (
natka813 [3]

Answer:

The flux (volume of water per unit time) through the hoop will also double.

Explanation:

The flux = volume of water per unit time = flow rate of water through the hoop.

The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.

This means that

Flow rate = AV

where A is the area of the hoop

V is the velocity of the water through the hoop

This flow rate = volume of water per unit time = Δv/Δt =Q

From all the above statements, we can say

Q = AV

From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2

3 0
4 years ago
What is a business cycle?
Anestetic [448]

Answer:

a cycle or series of cycles of economic expansion and contraction.

Explanation:

5 0
3 years ago
Information such as tolerances and scale can be found in the _______________ of an engineering drawing.
nasty-shy [4]

Answer:

Information such as tolerance and scale can be found in the <u>title block</u> of an engineering drawing

Explanation:

The title block of an engineering drawing can normally be found on the lower right and corner of an engineering drawing and it carries the information that are used to specify details that are specific the drawing including, the name of the project, the name of the designer, the name of the client, the sheet number, the drawing tolerance, the scale, the issue date, and other relevant information, required to link the drawing with the actual structure or item

8 0
3 years ago
A ramjet operates by taking in air at the inlet, providing fuel for combustion, and exhausting the hot air through the exit. Th
UNO [17]

Answer:

15300 N

Explanation:

\rho_i = Density of air at inlet

\dfrac{m}{t} = Mass flow rate = 60 kg/s

v_i = Inlet velocity = 225 m/s

\rho_o = Density of gas at outlet = 0.25\ \text{kg/m}^3

A_i = Inlet area

A_o = Outlet area = 0.5\ \text{m}^2

Since mass flow rate is the same in the inlet and outlet we have

\rho_iv_iA_i=\rho_ov_oA_o\\\Rightarrow v_o=\dfrac{\dfrac{m}{t}}{\rho_oA_o}\\\Rightarrow v_o=\dfrac{60}{0.25\times 0.5}\\\Rightarrow v_o=480\ \text{m/s}

Thrust is given by

F=\dfrac{m}{t}(v_o-v_i)\\\Rightarrow F=60\times (480-225)\\\Rightarrow F=15300\ \text{N}

The thrust generated is 15300 N.

8 0
3 years ago
Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12
Katarina [22]

Answer:

\dot W_{out} = 3374.289\,\frac{BTU}{s}

Explanation:

The model for the turbine is given by the First Law of Thermodynamics:

- \dot W_{out} + \dot m \cdot (h_{in} - h_{out}) = 0

The turbine power output is:

\dot W_{out} = \dot m\cdot (h_{in}-h_{out})

The volumetric flow is:

\dot V = \frac{\pi}{4} \cdot \left( \frac{2}{12}\,ft \right)^{2}\cdot (620\,\frac{ft}{s} )

\dot V \approx 13.526\,\frac{ft^{3}}{s}

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

\nu = 1.33490\,\frac{ft^{3}}{lbm}

The mass flow is:

\dot m = \frac{\dot V}{\nu}

\dot m = \frac{13.526\,\frac{ft^{3}}{s} }{1.33490\,\frac{ft^{3}}{lbm} }

\dot m = 10.133\,\frac{lbm}{s}

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

h = 1479.74\,\frac{BTU}{lbm}

State 2 (Saturated Vapor)

h = 1146.1\,\frac{BTU}{lbm}

The turbine power output is:

\dot W_{out} = (10.133\,\frac{lbm}{s} )\cdot (1479.1\,\frac{BTU}{lbm}-1146.1\,\frac{BTU}{lbm})

\dot W_{out} = 3374.289\,\frac{BTU}{s}

6 0
3 years ago
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