A pawn shop dealer has just purchased a ring that he was told was 24K Gold, which is pure Gold-197. He happens to have a mass spectrometer for just these occasions. He scraped a tiny sample off the ring and analyzed it in the mass spectrometer. The following spectra was generated. Determine if this was 24K Gold. If it was not, indicate the other metal(s) present, and the specific isotope of that element. Back up your answer using the information provided in the spectra and the periodic table. 100 Relative abundance (96) 50 o 65 107 197 Mass-to-charge ratio

Answer:

True!! took the test

Explanation:

The ball of mass m is dropped from a height h0 above the ground and falls to a height h1. The change in height can be expressed as h1 - h0. The mass of the ball, m, is a constant in this scenario and does not affect the distance the ball falls. The distance the ball falls is determined by the height it is dropped from (h0) and the gravitational force of the earth. Therefore, the distance the ball falls can be expressed as d = h1 - h0 = 1/2 gt^2, where g is the acceleration due to gravity and t is the time it takes for the ball to fall. The variables in this scenario are the initial height (h0), the final height (h1), and acceleration due to gravity.

To know more about force click here:

brainly.com/question/31932087

#SPJ11

The region of convergence corresponding to G(s) will be To = -1 and A = 1

Based on the information given, it should be noted that the value when we evaluate X(s) and specify its region of convergence will be:

X(s( = e^(5 + 5)/(5+5)

= -5.

The region where the function exists in the transfer function's pole/zero plot is known as the Region of Convergence. We prefer to deal with rational functions for the purpose of practical filter design, which may be defined by two polynomials, one for identifying the poles and the other for calculating the zeros, respectively.

The values of the finite numbers A and to such that the Laplace transform G(s) of g(t) = Ae-51 u(-t-to) has the same algebraic form as X(s) will be - 1 and the region of convergence corresponding to G(s) is 1.

Therefore based on the information, the region of convergence corresponding to G(s) will be To = -1 and A = 1.

Learn more about convergence on:

brainly.com/question/1521191

#SPJ1

Answer:

True

Explanation:

they put it through a process to be able to reuse it

Answer:

24cm sureeeereereeeeee

Focus primarily on highly precise specifics of what Cornell offers and how it matches with your interests and values. You can regard this as a "Why us?" essay with some optional "Why major" spice.

To learn more about essay refer :

https://brainly.com/question/94727

#SPJ4

The flow rates of the different components are :Flow rate of ethylene = 0.033 lb/hr (approx)Flow rate of oxygen = 0.038 lb/hr (approx)Flow rate of ethylene oxide = 2000 lb/hrFlow rate of CO2 and H2O = 0.013 lb/hr (approx).

Given dataImmediate feed to the reactor contains 3 moles of ethylene per mole of oxygenSingle-pass conversion of ethylene is 20%, and for every 100 moles of ethylene consumed in the reactor, 90 moles of ethylene oxide emerge in the reactor products.Flow rate of ethylene oxide = 2000 lb/hrSteps to determine all flow rates:We have to start by determining the number of moles of ethylene oxide to be produced per hour.Conversion of ethylene in a single pass = 20%Thus, the number of moles of ethylene converted per hour = (20/100) × 3 = 0.6 mol/hr.

Therefore, the number of moles of ethylene oxide produced per hour = (0.6 mol/hr) × (90/100) = 0.54 mol/hrFor every mole of ethylene oxide produced, 2 moles of ethylene are consumed and 1 mole of oxygen is consumed.Thus, the total number of moles of ethylene consumed per hour = (2/1) × 0.54 = 1.08 mol/hrThe number of moles of oxygen consumed per hour = 0.54 mol/hrLet us now convert these moles into mass flow rates using the molar mass of each component.

Flow rate of ethylene = (1.08 mol/hr) × (28.05 g/mol) × (1 lb/454 g) = 0.066 lb/hr (approx)Flow rate of oxygen = (0.54 mol/hr) × (32 g/mol) × (1 lb/454 g) = 0.021 lb/hr (approx)Flow rate of ethylene oxide = 2000 lb/hrFlow rate of CO2 and H2O = Sum of the flow rates of ethylene and oxygen - flow rate of ethylene oxide= (0.066 + 0.021) - (2000/454) = -4.4 lb/hrThis negative flow rate of CO2 and H2O indicates that there is an error in our calculations. We made an assumption that for every mole of ethylene oxide produced, 2 moles of ethylene are consumed.

However, this is not true, as the stoichiometry of the reaction shows that for every mole of ethylene oxide produced, 1 mole of oxygen is consumed. Therefore, we need to recalculate the flow rates using this stoichiometry.Flow rate of ethylene = (0.54 mol/hr) × (28.05 g/mol) × (1 lb/454 g) = 0.033 lb/hr (approx)Flow rate of oxygen = (0.54 mol/hr) × (32 g/mol) × (1 lb/454 g) = 0.038 lb/hr (approx)Flow rate of ethylene oxide = 2000 lb/hrFlow rate of CO2 and H2O = Sum of the flow rates of ethylene and oxygen - flow rate of ethylene oxide= (0.033 + 0.038) - (2000/454) = 0.013 lb/hr (approx).

Learn more about ethylene

https://brainly.com/question/14797464

#SPJ11

To implement a binary full adder with a dual 4-input multiplexer and a single inverter, you can use the following steps:

1. First, you need to understand the basic concept of a full adder. A full adder is a combinational logic circuit that adds three inputs, namely two binary digits (A and B) and a carry input (Cin) to produce two outputs, a sum output (S) and a carry output (Cout).

2. To implement a full adder using a dual 4-input multiplexer and a single inverter, you can use the following logic:

- Connect the A and B inputs to the select inputs of one of the multiplexers, and connect the Cin input to the select inputs of the other multiplexer.

- Connect the four possible combinations of A and B (00, 01, 10, 11) to the four data inputs of the first multiplexer.

- Connect the four possible combinations of Cin (00, 01, 10, 11) to the four data inputs of the second multiplexer.

- Connect the inverted output of the second multiplexer to the select input of the first multiplexer.

- Connect the output of the first multiplexer to the sum output (S) of the full adder.

- Connect the output of the second multiplexer to the carry input (Cin) of the next full adder in a cascaded full adder system.

3. This logic works because the dual 4-input multiplexer selects the appropriate input based on the values of A, B, and Cin, and the inverted output of the second multiplexer is used to complement the inputs to the first multiplexer, which produces the sum output (S) of the full adder.

4. Therefore, by using a dual 4-input multiplexer and a single inverter, you can implement a binary full adder in a simple and efficient way.

Learn more about binary digits here:

brainly.com/question/31931747

#SPJ11

Answer:

They moved fresh water around their vast empire with aqueducts and canals.

Explanation:

To anticipate health hazards in each of the eight areas of the AAP plant, the following information could be used:

1. Shipping/Receiving:

2. Hydraulic Presses:

3. Metal Working Lines:

4. Robotic Welding Stations:

5. Hand-Welding Stations in Rework Areas:

6. Small Paint Booths:

7. QA/QC Laboratory:

8. Final Inspection Area:

To anticipate health hazards in each area of Acme Automotive Parts (AAP) manufacturing facility, the following information would be relevant.

1. Shipping/Receiving: Material Safety Data Sheets (MSDS) for chemicals received, potential for heavy lifting injuries, and risks associated with forklift operations.

2. Hydraulic Presses and Metal Working Lines: MSDS for lubricants and coolants, risks of crush injuries, and exposure to metal fumes.

3. Robotic and Hand-Welding Stations: MSDS for welding materials, welding fume exposure, and risk of burns.

4. Paint Booths: MSDS for paints and coatings, potential for volatile organic compounds (VOCs) exposure, and respiratory hazards.

5. QA/QC Laboratory: MSDS for chemicals used in testing, potential for chemical exposure, and ergonomic risks.

6. Final Inspection Area: Ergonomic risks associated with repetitive tasks, MSDS for any chemicals used, and general safety precautions.

Learn more about Acme Automotive Parts  at:

https://brainly.com/question/30694117

#SPJ2

The smallest allowable depth d of the milled portion of the bar is 4 mm.

To determine the smallest allowable depth d of the milled portion of the bar, we need to consider the applied force (P) and the maximum allowable stress σ(all). The formula for stress in a rectangular cross-section is:
σ = P / A
where σ is the stress, P is the applied force, and A is the area of the cross-section. In this case, we are given that P = 18 kN and σ(all) = 135 MPa.
First, convert the units of P and σ(all) to be consistent:
P = 18 kN = 18,000 N
σ (all) = 135 MPa = 135 x 10⁶ N/m²
Next, we need to find the area of the milled portion of the bar. Since it has a square cross-section, the area A can be expressed as:
A = a * d
where a = 30 mm = 0.03 m and d is the depth of the milled portion in meters.
Now, we can use the stress formula to find the smallest allowable depth d:
σ(all) = P / A
135 x 10⁶ N/m² = 18,000 N / (0.03 m * d)
to solve for d, divide both sides by 0.03 m:
(135 x 10⁶ N/m²) / (0.03 m) = 18,000 N / d
Rearrange the equation to isolate d:
d = 18,000 N / [(135 x 10⁶ N/m²) * (0.03 m)]
Solve for d:
d = 0.004 m = 4 mm
So, the smallest allowable depth d of the milled portion of the bar is 4 mm.

Learn more about "stress " at: https://brainly.com/question/30369058

#SPJ11

yes it will

Explanation:

The algorithm determined the greatest number based on the provided information first, and after four iterations, the array is [2, 4, 5, 7, 8, 1, 3]. Hence it stands to reason that the sorting algorithm.

With a selection sort, the typical number of comparisons for each pass will always be half the number of items to be sorted. There are 36 comparisons for eight items, or 1/2(82 + 8) = 1/2(64 + 8) = 1/2(72).

Sorting the array and then looking for the second member from the sorted array that is not equal to the largest element can be used as a brute force method to locate the second largest element in an array. The brute approach has an O(n log(n)) time complexity, where n is the number of entries in the array.

to know  more about algorithms here:

brainly.com/question/22984934

#SPJ4

For shop spllls larger than 12 gallons or 100 pounds, contact the County where your shop is located. Option a is the right choice.

The U.S. Coast Guard staffs the National Response Centre (NRC), a component of the officially established National Response System, around-the-clock.

All oil, synthetic, radioactive, organic, and etiological discharges into the atmosphere anywhere in the United States and its territories are announced using this government site.

The NRC also investigates reports of questionable behavior and security lapses in the territorial waters of the United States and its regions. The national response center has expert-level skills in Digital Forensics, Technical Investigation, and Information System Security Audits.

To learn more about the national response center

https://brainly.com/question/14398527

#SPJ4

The diameter of a tube or pipe can significantly impact the flow rate of a fluid. The relationship between tube diameter and flow rate is described by the principles of fluid dynamics, specifically the Hagen-Poiseuille equation for laminar flow. According to this equation, the flow rate (Q) is directly proportional to the fourth power of the tube radius (r) and inversely proportional to the viscosity of the fluid (η) and the length of the tube (L).

Q ∝ r^4 / (ηL)

Based on this equation, the following recommendations can be made regarding the effect of different tube sizes on flow rate:

Larger Diameter, Higher Flow Rate: Increasing the diameter of a tube will result in a higher flow rate, assuming other factors remain constant. This is because the flow rate is directly proportional to the fourth power of the tube radius. Therefore, a small increase in diameter can lead to a substantial increase in flow rate.

Smaller Diameter, Lower Flow Rate: Conversely, reducing the diameter of a tube will decrease the flow rate. The fourth power relationship means that even a slight reduction in diameter can result in a significant decrease in flow rate.

While larger tube diameters generally offer higher flow rates, they can also lead to increased pressure drop along the length of the tube. Pressure drop refers to the decrease in pressure as fluid flows through the tube due to friction with the tube walls. It's important to consider the acceptable pressure drop for your system, as excessively high pressure drop can lead to inefficiencies or practical issues.

Learn more about viscosity on:

https://brainly.com/question/30759211

#SPJ1

Tires are an essential component of any vehicle, and their maintenance is critical for ensuring safety, performance, and longevity. One of the factors that significantly impact tire pressure is temperature.

As the temperature changes, so does the air pressure inside the tire. In general, tires typically gain or lose about 1 psi for each ten-degree Fahrenheit change in temperature. For example, if the temperature drops by 20 degrees, the tire pressure may decrease by about 2 psi. Similarly, if the temperature increases by 10 degrees, the tire pressure may increase by 1 psi.

It's crucial to note that tire pressure is vital for the overall safety and performance of the vehicle. Low tire pressure can cause several problems, including reduced fuel efficiency, decreased handling and stability, and increased wear and tear on the tires. Therefore, it's essential to regularly check and maintain the tire pressure, especially during temperature fluctuations.

In conclusion, understanding the impact of temperature on tire pressure is crucial for maintaining tire health and ensuring optimal performance and safety. Tires typically gain or lose about 1 psi for each ten-degree Fahrenheit change in temperature, and regularly monitoring tire pressure is vital for a safe and smooth driving experience.

Know more about air pressure here:

https://brainly.com/question/15189000

#SPJ11

To calculate the average memory access time for instruction accesses (a), data reads (b), data writes (c), and the overall CPI including memory accesses (d), we need to consider the cache hierarchy and memory system parameters given.

a. Average Memory Access Time for Instruction Accesses:

The instruction cache (I-cache) is direct-mapped with a 2% miss rate and 32-byte blocks. The I-cache imposes no penalty on hits.

Average memory access time for instruction accesses = Hit time + Miss rate * Miss penalty

Given:

Hit time = 0 (no penalty on hits)

Miss rate = 2% = 0.02

Miss penalty = Access time of L2 cache = 15 ns

Average memory access time for instruction accesses = 0 + 0.02 * 15 ns = 0.3 ns

b. Average Memory Access Time for Data Reads:

The data cache (D-cache) is direct-mapped with a 5% miss rate and 16-byte blocks. The D-cache is write-through, but there is a write buffer that eliminates stalls for 95% of all writes.

Average memory access time for data reads = Hit time + Miss rate * Miss penalty

Given:

Hit time = 0 (no penalty on hits)

Miss rate = 5% = 0.05

Miss penalty = Access time of L2 cache = 15 ns

Average memory access time for data reads = 0 + 0.05 * 15 ns = 0.75 ns

c. Average Memory Access Time for Data Writes:

For data writes, there is a write buffer on the D-cache that eliminates stalls for 95% of all writes. The write buffer avoids the need to access the L2 cache for most writes.

Average memory access time for data writes = Hit time + (1 - Write buffer hit rate) * Miss penalty

Given:

Hit time = 0 (no penalty on hits)

Write buffer hit rate = 95% = 0.95

Miss penalty = Access time of L2 cache = 15 ns

Average memory access time for data writes = 0 + (1 - 0.95) * 15 ns = 0.75 ns

d. Overall CPI including Memory Accesses:

To calculate the overall CPI including memory accesses, we need to consider the fraction of memory references that cause cache misses and access the main memory.

Overall CPI = CPI (excluding memory accesses) + (Memory access time / Clock cycle time)

Given:

CPI (excluding memory accesses) = 1.35

Memory access time = Average memory access time for instruction accesses + (Memory references causing cache misses * Average memory access time for data reads) + (Memory references causing cache misses * Average memory access time for data writes)

Clock cycle time = 1 / (Processor frequency)

Memory references causing cache misses = Instruction references * Instruction miss rate + Data references * Data miss rate

Instruction references = 20% of all instructions

Data references = 10% of all instructions

Calculating the values:

Memory references causing cache misses = (20% * 0.02) + (10% * 0.05) = 0.006

Memory access time = 0.3 ns + (0.006 * 0.75 ns) + (0.006 * 0.75 ns) = 0.3045 ns

Clock cycle time = 1 / (1.1 GHz) = 0.909 ns

Overall CPI including Memory Accesses = 1.35 + (0.3045 ns / 0.909 ns) = 1.35 + 0.335 = 1.685

Therefore:

a. Average memory access time

Learn more about Average memory access time from

https://brainly.com/question/31978184

#SPJ11

The power delivered by the four-cylinder, two-stroke 2.4-L diesel engine operating on an ideal Diesel cycle with given parameters can be calculated using the cold-air-standard assumptions.

To determine the power delivered by the engine, we can use the basic equation for power in a reciprocating engine:
Power = (Work per cycle) * (Number of cycles per unit time)
In the case of the Diesel cycle, the work per cycle can be expressed as:
Work per cycle = (Heat added at constant pressure) - (Heat rejected at constant volume)
Using the cold-air-standard assumptions, we can calculate the heat added and heat rejected per cycle based on the given parameters and properties of air. The heat added can be calculated using the equation:
Qin = cp * T1 * [(r^k) - 1]
where cp is the specific heat at constant pressure, T1 is the initial temperature, r is the compression ratio, and k is the specific heat ratio.
Similarly, the heat rejected can be calculated using the equation:
Qout = cv * T3 * [(r^k) - 1]
Where cv is the specific heat at constant volume, T3 is the temperature at the end of the expansion process (given as 70°C), and r is the cutoff ratio.
Once the heat added and heat rejected per cycle are determined, the work per cycle can be calculated as the difference between the two. Finally, multiplying the work per cycle by the number of cycles per unit time (determined by the engine speed) gives the power delivered by the engine.
In this case, with the given parameters and properties of air, the calculated power delivered by the engine will be in kilowatts.

Learn more about diesel cycle here
https://brainly.com/question/13040349

 #SPJ11

Try to be as relax as possible.

Provide names of all parties involved.

Provide vehicle information and identification details.

Provide full names, address, registration numbers and insurance company details.

Explanation:

After a collision one may be confused, afraid and have no attention about the details that what happened because all the collision event happens in a short interval of time. So the first thing one should do during information exchange is to sit back and relax and be calm so that one can remind the things at some extent. After that provide all the details about injured people and the involved vehicles.

The entropy of 2 moles of and ideal gas expanding freely to 3 times it's initial volume is 18.3J/k

The following formula is required:

∆S = nx R x ℓn x (Vf/Vi)

Where

n = number of moles of gas (n = 2)

R = gas constant (R = 8.314 J/(mol * K))
Vf = final volume (Vf = 3.V1)

Vi = intial volume

Vi = 1L (Asumption )

∆S = 2 x 8.314 x 1.099

∆S =18.3 j/K

Learn more about entropy:
https://brainly.com/question/13135498
#SPJ4

Translation:

Calculate the entropy of 2 moles of an ideal gas expanding freely to three times its initial volume, use: ∆S =n・R・ℓn (Vf / Vi)

To convert the pressure reading from psi to inches of mercury, we can use the following equation:

P_hg = P_gage * (H_hg / H2O)

where P_hg is the pressure in inches of mercury, P_gage is the gage pressure reading in psi, H_hg is the density of mercury (13.6 g/cm^3), and H2O is the density of water (1 g/cm^3).

Plugging in the values, we get:

P_hg = 40 psi * (13.6 g/cm^3 / 1 g/cm^3) / (2.54 cm/in)

Simplifying, we get:

P_hg = 547.2 / 2.54 in

P_hg ≈ 215.35 inHg

Therefore, if a mercury manometer replaces the gage, the reading would be approximately 215.35 inches of mercury.

Learn more about   psi to inches of mercury from

https://brainly.com/question/31666388

#SPJ11

Answer:

The answer is below

Explanation:

Case hardening is a form of steel hardening that is applied on mild steel with a high temperature of heat.

It results in material forming a hard surface membrane, while the inner layer is soft.

It is mostly used for universal joints, construction cranes, machine tools, etc.

On the other hand, Through hardening is a form of steel hardening in engineering that involves heat treatment of carbon steel.

It increases the hardness and brittleness of the material.

It is often used for axles, blades, nuts and bolts, nails, etc.

Answer:

\(W = 12.5\ lb/ft^2\) --- in \(lb/ft^2\)

\(W = 598.503N/m^2\) --- in \(N/m^2\)

Explanation:

Given

\(Weight=2000lb\)

\(Area = 160ft^2\)

Required

Determine the wing loading (W)

Wing loading is calculated using:

\(W = \frac{Weight}{Area}\)

Substitute values for Weight and Area

\(W = \frac{2000lb}{160ft^2}\)

\(W = 12.5\ lb/ft^2\)

The answer can also be converted to N/m^2

\(1\ lb = 4.4482216153N\)

\(1ft^2 = 0.092903m^2\)

So, we have:

\(W = \frac{12.5 * 4.4482216153N}{0.092903m^2}\)

\(W = \frac{55.6027701912N}{0.092903m^2}\)

\(W = 598.503N/m^2\)

Hence, the wing loading is:

\(W = 12.5\ lb/ft^2\) --- in \(lb/ft^2\)

\(W = 598.503N/m^2\) --- in \(N/m^2\)

The number of paths in the network is six.

The paths are:
C - A - T - N - EndC - A - T - R - N - EndC - B - P - U - N - EndC - B - P - U - T - R - N - EndC - B - P - A - T - N - EndC - B - P - A - T - R - N - End
2. The critical path is C - A - T - N - End with a duration of 24 weeks.

3. The float on activity U is 1 week.

4. If activity B takes three weeks longer than planned, it will affect the duration of path 3 and the entire project. It will elongate the project duration to 29 weeks.

2. b) The critical path is A-C-E-G with a duration of 18 weeks.

(c) The normal expected project completion time is 18 weeks.

(d) The total project cost using normal time is RM 7600.

(e) To complete the project two weeks faster, activities C and E can be crashed for the minimum cost.

The additional cost incurred will be RM 600.

The total cost is RM 8200.

3 The inventory cost that will minimize the total cost of inventory is obtained using the formula,

TAC = IC + OC + HCS

Where TAC

= Total Annual CostIC = Inventory CostOC = Ordering CostHCS = Holding and Carrying Cost.

The probability distribution is shown below. Demand 0 1 2 3 4 5Probability 0.20 0.15 0.25 0.20 0.10 0.10

To determine the optimal order quantity and reorder policy that would minimize total cost,

the EOQ (Economic Order Quantity) model will be used.

The EOQ can be determined using the formula,

Q = √((2DCO)/CH')

Where Q = Economic Order Quantity

D = Demand'

C0 = Cost per order

H = Holding cost

C = IC per unit

Let’s assume the cost per unit, C = RM 500,

the holding cost, H = 0.2,

and demand, D = 3 units per day.

Hence,CO = RM 800 (C0 = Cost per order, which is the same as ordering cost)

Then,Q = √((2 x 3 x RM 800)/RM 100)Q = 24 units per order

Then, the reorder level can be determined using the formula,

ROL = d * LWhere L = Lead time

ROL = 3 x 5 = 15 units

The total annual cost can be determined by substituting the values into the formula,

TAC = IC + OC + HCS

where IC = 0.5 x RM 500

= RM 250OC

= (365/EOQ) x C0OC

= (365/24) x RM 800OC

= RM 12133.33HCS

= (Q/2) x H x dHCS

= (24/2) x 0.2 x 3HCS

= RM 7.20

TAC = RM 250 + RM 12133.33 + RM 7.20TAC = RM 12390.53

Hence, the order quantity that will minimize the total cost of inventory is 24 units.

To know more about paths visit:

https://brainly.com/question/31522531

#SPJ11

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

36

1

​

P(X=3)=P(1,2)+P(2,1)=

36

2

​

=

18

1

​

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

36

3

​

=

12

1

​

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

36

4

​

=

9

1

​

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

36

5

​

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

36

6

​

=

6

1

​

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

36

5

​

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

36

4

​

=

9

1

​

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

36

3

​

=

12

1

​

P(X=11)=P(5,6)+P(6,5)=

36

2

​

=

18

1

​

P(X=12)=P(6,6)=

36

1

​

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

i

​

⋅P(X

i

​

)

=2×

36

1

​

+3×

18

1

​

+4×

12

1

​

+5×

9

1

​

+6×

36

5

​

+7×

6

1

​

+8×

36

5

​

+9×

9

1

​

+10×

12

1

​

+11×

18

1

​

+12×

36

1

​

=

18

1

​

+

6

1

​

+

3

1

​

+

9

5

​

+

6

5

​

+

6

7

​

+

9

10

​

+1+

6

5

​

+

18

11

​

+

3

1

​

=7

E(X

2

)=∑X

i

2

​

⋅P(X

i

​

)

=4×

36

1

​

+9×

18

1

​

+16×

12

1

​

+25×

9

1

​

+36×

36

5

​

+49×

6

1

​

+64×

36

5

​

+81×

9

1

​

+100×

12

1

​

+121×

18

1

​

+144×

36

1

​

=

9

1

​

+

2

1

​

+

3

4

​

+

9

25

​

+5+

6

49

​

+

9

80

​

+9+

3

25

​

+

18

121

​

+4

=

18

987

​

=

6

329

​

=54.833

Then, Var(X)=E(X

2

)−[E(X)]

2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

​

=

5.833

​

=2.415

To determine the minimum bit rate of an ADC (Analog-to-Digital Converter) with a 1000-level quantizer and a bandwidth of 20 kHz, the minimum bit rate from this ADC is 400 kHz.

In this case, the signal has a bandwidth of 20 kHz, so the minimum sampling rate required is 2 times the bandwidth, which is 2 * 20 kHz = 40 kHz. The minimum sampling rate corresponds to the minimum bit rate.

To convert an analogue signal with a 20 kHz bandwidth to a binary format using a 1000-level quantizer, each level of the quantizer requires a certain number of bits. Since there are 1000 levels, we need at least log2(1000) bits to represent each level. Rounded up to the nearest integer, log2(1000) is 10.

Therefore, the minimum bit rate of the ADC is the product of the minimum sampling rate and the number of bits per sample:

Minimum bit rate = Minimum sampling rate * Number of bits per sample

                = 40 kHz * 10 bits

                = 400 kHz

Hence, the minimum bit rate from this ADC is 400 kHz.

Learn more about analogue signal here: https://brainly.com/question/33183802

#SPJ11

The design of the input shaft includes gear, bearings, key, retaining rings, and shaft specifications. It involves selecting appropriate components and determining their specifications for efficient operation.

Designing the input shaft involves careful consideration of various factors to ensure efficient and reliable operation. The gear, bearings, key, retaining rings, and shaft specifications are critical components in this process. The gear selection is based on factors such as torque requirements, speed, and desired gear ratio.  The bearings must be chosen to handle the expected loads and provide smooth rotation.

The key and retaining rings ensure proper alignment and secure attachment of the gear to the shaft. The shaft specification includes determining its material, dimensions, and surface finish to meet strength, stiffness, and durability requirements. Factors like torque, speed, and operating conditions play a crucial role in selecting the appropriate material and ensuring the shaft can withstand the applied forces.

Careful consideration of these specifications and component choices ensures optimal performance and reliability of the input shaft in the specific application.

Learn more about input shaft

brainly.com/question/31974012

#SPJ11