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liq
1 month ago
6

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

hography and the resulting ability to print ever-smaller features. Modern circuits are made using a variety of complicated lithography techniques, with the goal to make electronic traces as small and as close to each other as possible (to reduce the overall size, and thus increase the speed). In the end, though, all-optical techniques are limited by diffraction.
Assume we have a scannable laser that draws a line on a circuit board (the light exposes a line of photoresist, which then becomes impervious to a subsequent chemical etch, leaving only the narrow metal line under the exposed photoresist). Assume the laser wavelength is 248.0 nm (Krypton Fluoride excimer laser), the initial beam diameter is 1.0 cm, and the focusing lens (diameter = 1.3 cm) is extremely 'fast', with a focal length of only 0.625 cm.
a. What is the approximate width w of the line?
b. What is the minimum resolvable line separation between adjacent lines?
c. If the laser wavelength is instead reduced to 157 nm, what is the new minimum resolvable line separation?
Physics
1 answer:
Maru [3.3K]1 month ago
6 0
0.000003782 m 0.000001891 m 0.000001197125 m Explanation: Wavelength = 248 nm D = Beam diameter = 1 cm f = Focal length = 0.625 cm. The angle can be expressed as The width is defined as The required width is determined to be 0.000003782 m. The minimum resolvable line separation stands at 0.000001891 m when adjustments are made. The new minimum resolvable line separation, assuming a wavelength reduction to 157 nm, is to be determined.
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Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Ostrovityanka [3204]
This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
2 months ago
Read 2 more answers
Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,
Sav [3153]

Answer:

All three pendulums will have the same angular frequencies.

Explanation:

For a simple pendulum, the time period using the approximation sin(\theta )\approx \thetais expressed as:

T=2\pi\sqrt{\frac{l}{g}}

The angular frequency \omega is defined as

\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}

Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.

6 0
2 months ago
A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
ValentinkaMS [3465]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is \frac{4}{3} \pi (\frac{D}{2})^3

Thus \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

The radius of the disk is 1.15(\frac{ D}{2} )

Applying angular momentum conservation;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disk = m*\frac{ \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
2 months ago
The amount of electric energy consumed by a 60.0-watt lightbulb for 1.00 minute could lift a
Sav [3153]

Answer:

Explanation:

For a 60W light bulb used for 1 minute:

P = 60 W

t = 1 minute = 60 seconds

This energy is capable of lifting an object weighing 10N.

W = 10N

This indicates conversion of electrical energy into potential energy.

Let's calculate the electrical energy:

Power describes the rate of work done.

Power = Work / time

Thus, work = power × time

Work = 60 × 60

Work = 3600 J

Potential energy calculation:

P.E = mgh

Where the weight is given by:

W = mg

Therefore, P.E = W·h

P.E = 10·h

Thus, we equate:

Potential energy = Electrical energy

P.E = Work

10·h = 3600

Dividing both sides by 10 gives:

h = 3600 / 10

h = 360m

The object can be lifted to a height of 360m.

6 0
2 months ago
Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl
serg [3582]

Answer:

b) TA = TB = TC

Explanation:

  • When the blocks are brought into contact and isolated from the environment, they will exchange heat until they achieve thermal equilibrium.
  • During this exchange, the hotter body will lose heat, which will be gained by the cooler body.
  • The equilibrium state will be established once this equation is satisfied:

       \Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )

  • Substituting the initial temperatures T₀A = 300º C and T₀B = 400ºC, while simplifying for equal block masses mA = mB, enables us to solve for the final temperature, Tfin:

       (400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC

  • At equilibrium, when both blocks combine, they will yield a uniform final temperature of 350ºC.
  • When block C, also at this temperature, makes contact, all three blocks will simultaneously reflect this final temperature of 350 ºC.
  • Therefore, option b) is correct.
8 0
2 months ago
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