One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit
1 answer:
Answer:
0.000003782 m
0.000001891 m
0.000001197125 m
Explanation:
= Wavelength = 248 nm
D = Diameter of beam = 1 cm
f = Focal length = 0.625 cm
The angle is given by
The width is given by
The required width is 0.000003782 m
Minimum resolvable line separation is given by
The minimum resolvable line separation between adjacent lines is 0.000001891 m
when
The new minimum resolvable line separation between adjacent lines is
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Answer:
The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°
Explanation:
In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).
So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.
After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:
we can see there are only two forces in x, which are the weight on x and the static friction, so:
when solving for the static friction we get:
We know the weight is found by multiplying the mass by the acceleration of gravity, so:
W=mg
and:
we can substitute this on our sum of forces equation:
the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:
so we can substitute this on our equation:
but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.
In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:
when solving for N we get:
When seeing the free-body diagram we can determine that:
so we can substitute that in the sum of y-forces equation, so we get:
we can go ahead and substitute this equation in the sum of forces in x equation so we get:
we can divide both sides of the equation into mg so we get:
as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for we get:
when simplifying this we get that:
now we can solve for the angle so we get:
and we can substitute the given value so we get:
which yields:
α=19.29°
which rounds to:
α=19°
Answer:
A) E = 145.6 N / C , B) y= 2,8 10-7 m with a downward direction
C) he shape of the trajectory of the two particles is to simulate a parabola,
D) F_{e} /F_{g} = 10³⁴
Explanation:
A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric
F = m a
- e E = m a
a = - e E / m
with the field directed downward, the acceleration is in the vertical upward direction.
We look for how much the electron moves with kinematics, in the x direction there is no acceleration,
x axis (parallel to plates)
x = v₀ t
t = x / v₀
y axis (perpendicular to plates)
y = y₀ + t + ½ a t²
Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known
y = ½ a t²
we substitute
y = ½ (e E /m) (x / v₀)²
y = ½ e x2 /m v₀² E
we look for the electric field
E = 2 m y v₀² / e x²
where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m
let's calculate
E = 2 9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)
E = 145.6 N / C
B) The electron is exchanged for a proton
Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates
y = ½ e x² / m v₀² E
y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)² 145.6
y = 28.4375 10⁻⁸ m
since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small
In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)
C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards
D) The force of gravity
= G m M / R²
electron
Between the electron and the positive charges of the conducting plate
F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²
F_{g} = 4.1 10⁻⁵¹ N
electric force
F_{e} = -e E
F_{e} = - 1.6 10⁻¹⁹ 145.6
F_{e} = 2.3 10⁻¹⁷ N
let's look for the reason between these two forces
F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹
F_{e} /F_{g} = 10³⁴
We see that the electric force is many orders of magnitude higher than the gravitational force.