A thin-film gold conductor for an integrated circuit in a satellite application is deposited from a vapor, and the deposited gol

Answer:

Refer to the attached document

1512 ft

Explanation:

Because the acceleration is constant or zero, the acceleration-time graph consists of horizontal segments. The values for t2 and a4 are derived as follows:

Acceleration - Time

0 < t < 6: Velocity change = area beneath the a–t graph

V_6 - 0 =  (6 s)(4 ft/s²) = 24 ft/s

6 < t < t2: The velocity rises from 24 to 48 ft/s,

Velocity change = area beneath the a–t graph

48 - 24 = (t2 - 6) * 6

t2 = 10 s

t2 < t < 34: The velocity remains constant, meaning acceleration is zero.

34 < t < 40: Velocity change = area beneath the a–t graph

0 - 42 = 6*a4

a4 = - 8 ft / s²

A negative acceleration shows the area lies below the t-axis, indicating a decrease in speed.

Velocity - Time

Since acceleration remains constant or zero, the v−t graph is made up of linear segments connecting the calculated points.

Position change = area beneath the v−t graph

0 < t < 6:  x6 - 0 = 0.5*6*24 = 72 ft

6 < t < 10:    x10 - x6 = 0.5*4*(24 + 48) = 144 ft

10 < t < 34: x34 - x10 = 48*24 = 1152 ft

34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft

Summing these position changes yields the distance from A to B:

d = x40 - 0 = 1512 ft

Answer:

Ps=19.62N

Explanation:

A thorough explanation of the answer can be found in the attached files.

Answer:

Time constant = 15.34 seconds

The thermometer indicates an error margin of 0.838°

Explanation:

Given

t = 1 minute = 60 seconds

c(t) = 98% = 0.98

According to the details provided, the thermometer functions as a first-order system.

The transfer function of such a system is expressed as;

C(s)/R(s) = 1/(sT + 1).

To determine the time constant, the step response must be evaluated.

This is defined as

r(t) = u(t) --- Applying Laplace Transformation

R(s) = 1/s

Replacing 1/s back into C(s)/R(s) = 1/(sT + 1).

What we have

C(s)/1/s = 1/(sT + 1)

C(s) = 1/(sT + 1) * 1/s

C(s) = 1/s - 1/(s + 1/T) --- Taking Inverse Laplace Transformation

L^-1(C(s)) = L^-1(1/s - 1/(s + 1/T))

Given that e^-t <–> 1/(s + 1) --- {L}

1 <–> 1/s {L}

Thus, the unit response c(t) = 1 - e^-(t/T)

Substituting 0.98 for c(t) and 60 for t

0.98 = 1 - e^-(60/T)

0.98 - 1 = - e^-(60/T)

-0.02 = - e^-(60/T)

e^-(60/T) = 0.02

ln(e^-(60/T)) = ln(0.02)

-60/T = -3.912

T = -60/-3.912

T = 15.34 seconds

It follows that the time constant = 15.34 seconds

The error signal is characterized by

E(s) = R(s) - C(s)

Where the temperature shifts at 10°/min; which equals 10°/60 s = 1/6

Thus,

E(s) = R(s) - 1/6 C(s)

Calculating C(s)

C(s) = 1/s - 1/(s + 1/T)

C(s) = 1/s - 1/(s + 1/15.34)

Note that R(s) = 1/s

Thus, E(s) turns into

E(s) = 1/s - 1/6(1/s - 1/(s + 1/15.34))

E(s) = 1/s - 1/6(1/s - 1/(s + 0.0652)

E(s) = 1/s - 1/6s + 1/(6(s+0.0652))

E(s) = 5/6s + 1/(6(s+0.0652))

E(s) = 0.833/s + 1/(6(s+0.0652)) ---- Taking Inverse Laplace Transformation

e(t) = 1/6e^-0.652t + 0.833

In a first-order system, a steady state condition is reached when the time is four times the time constant.

Thus,

Time = 4 * 15.34

Time = 61.36 seconds

Consequently, e(t) becomes

e(t) = 1/6e^-0.652t + 0.833

e(t) = 1/(6e^-0.652(61.36)) + 0.833

e(t) = 0.83821342824942664566211

e(t) = 0.838 --- Rounded off

Thus, the thermometer reveals an error of 0.838°

Explanation:

At a temperature of and relative humidity of 86%, the humidity ratio stands at 0.0223 with a specific volume of 14.289.

At a temperature of and relative humidity of 40%, the humidity ratio is 0.0066 while the specific volume is 13.535.

To determine the mass of air, the following formula can be used:

Now, we will calculate the volume

The duration required to fill the cistern can be determined with the equation:

By substituting the values into the preceding formula, we find:

Thus, the hours necessary to fill the cistern amount to 4.65 hours.