water flows at 8 cms in a trapezoidal channel with a width of 2 m and side slopes of 3h:2v. over a distance of 170 m, the width expands to 2.7 m, with the side slopes remaining the same. if the flow depth is 1m at both sections and the channel slope is 0.001, calculate the head loss between the sections. what is power dissipated in kilowatts.

Answer:

TRUE

Explanation:

Answer:

first is the parentheses, (3+2)=5 next is the exponent 5^2=25, next is the division 5 / 5 = 1, then the multiplication 4*1=4 and then you add 4+25=29. so the answer is 29.

Answer:

chehdhfhfhd

Explanation:

fjrjshrhdhr

The basic requirement of measurements is to have a standard or reference point against which to compare the quantity being measured. This standard or reference point should be well-defined and stable, and the measurement process should be repeatable and consistent. Additionally, it is important to ensure that the measurement equipment is calibrated and in good working condition.

Answer:

The most basic requirement for measurements is the presence of a standard or reference point against which the quantity being measured can be compared. The measurement process should be repeatable and consistent, and the standard or reference point should be well-defined and stable. Furthermore, ensure that the measurement equipment is calibrated and in good working order.

Explanation:

Answer:

oa

Explanation:

Answer:

safety factor

Explanation:

i got it correct on the test

The electrons in the innermost shell has the lowest energy level while the electrons in the outermost shell has the highest energy level.

Coulombs of charge were moved when the car battery moved 4.75 × 1021 electrons through the starter motor.

Given, the car battery moves 4.75 × 1021 electrons through the starter motor.

1 electron = 1.6 × 10^-19 C \(1.6 × 10^-19 C\)

Charge moved by 4.75 × 1021

electrons= (4.75 × 1021) × (1.6 × 10^-19) C

\((4.75 × 1021) × (1.6 × 10^-19) C\)

= 7.6 C

7.6

For an electron to increase its energy level, it has to move to an higher shell; and to decrease its energy level it has to move to a lower shell.

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The frequency (in Hz) of the radiation released by the transition of an electron in a hydrogen atom from the n = 5 level to the n = 3 level is 7.68 x 10¹⁴ Hz

The frequency (in Hz) of the radiation released by the transition of an electron in a hydrogen atom can be calculated using the Rydberg formula:

1/λ = R (1/n1² - 1/n2²)

where λ is the wavelength of the radiation,

R is the Rydberg constant (3.29 x 10¹⁵ Hz), and

n1 and n2 are the initial and final energy levels of the electron in the hydrogen atom, respectively.

For the transition from n=5 to n=3, we have:

1/λ = R (1/3² - 1/5²)

1/λ = R (1/9 - 1/25)

1/λ = R (16/225)

λ = 225/16R

λ = 225/(16 x 3.29 x 10¹⁵)

λ ≈ 390.6 nm

The frequency of the radiation can be obtained by using the speed of light (c) and the wavelength (λ):

c = fλ

f = c/λ

f = (2.998 x 10⁸ m/s)/(390.6 x 10⁻⁹ m)

f ≈ 7.68 x 10¹⁴ Hz

Therefore, the frequency is approximately 7.68 x 10¹⁴ Hz.

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a. The power factor is 0989

b. The average power from the source is 2.125 kW

Power factor is the ratio of the real power absorbed by the load to the apparent power.

a. What is the power factor.

To find the power factor, we nedd to find the equivalent impedance of the circuit.

First, the the 8Ω resistor and -j6 Ω capacitor impedance are in series, their equivalent impedance is Z₁ = 8 Ω+ (- j6) Ω = (8 - j6)Ω.

Now, Z₁ is parallel to the Z₂ = j4 Ω impedance.

So, their equivalent inpedance is

Z₃ = Z₁Z₂/(Z₁ + Z₂)

= (8 - j6)j4 Ω/(8 - j6 + j4)

= (32 - j24)/(8 - j2)

Rationalizing the denominator, we have that

=  (32 - j24)/(8 - j2) × (8 + j2)/(8 + j2)

= [32 × 8 + 32 × j2 - j24 × 8 + (-j42 × -j2)]/(8² + 2)²

= [256 + j64 - 192j + j²84)]/(8² + 2²)

= [256 + j64 - 192j - 84)]/(64 + 4)

= (172 - j128)/68

= (43 - j32)/17

Since the 10 Ω resistor is in series with Z₃, the equivalent impedance is

Z₄ = 10 Ω + (43 - j32)/17 Ω

= (10 + 43/17 - j32/17 )Ω

= (170 + 43)/17 - j32/17

= 213/17 - j32/17

We know that for an impedance Z = a + jb , tanФ = b/a

So, for Z₄, tanФ = b/a

= -32/17 ÷ 213/17

= -32/213

since the trigonometric identity

tan²Ф + 1 = sec²Ф

secФ = ±√(tan²Ф + 1)

So, substituting tanФ into the equation, we have that

secФ = ±√(tan²Ф + 1)

secФ = ±√[(-32/213)² + 1)

= ±√[(1024 + 45369)/45369)]

= ±√[46393/45369)]

= ±215.39/213

Now, the power factor P.F = cosФ

Since secФ = 1/cosФ

cosФ = ±213/215.39

= ±0.989

Since tanФ is negative, Ф is in the fourth quadrant.

So, cosФ = 0.989

So, the power factor is 0989

b. The average power

The average power P = I²R where

Now, I = V/Z₄ where

We know that for an impedance Z = a + jb , tanФ = b/a

So, for Z₄ = , tanФ = b/a

= -32/17 ÷ 213/17

= -32/213

Ф = tan⁻¹(-32/213)

= -8.544

Also, the magnitude of impedance, Z = √(a² + b²)

So, the magnitude of Z = √[(213/17)² + (-32/17)²]

= √[(213)² + (-32)²]/17

= √[45369 + 1024]/17

= 215.39/17

= 12.67

So, Z = 12.67 ∠-8.544

So, the current I = V/Z

= 165 ∠ 0°/ 12.67 ∠-8.544

= 165/12.67 ∠ 0°- (-8.544)

= 13.02 ∠ 8.544° A

So, the average power P = I²R where

so, P = (13.02 A)² × 213/17 Ω

= 169.60 A² × 213/17 Ω

= 36123.754/17 A²Ω

= 2124.93 W

= 2.125 kW

The average power is 2.125 kW

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Answer:

its c

Explanation:

According to the question the Absolute velocity of B is V_B = (-2cosθ, 2sinθ) m/s.

Velocity is the rate at which an object changes its position in a given direction. It is a vector quantity, meaning it has both magnitude (the speed of movement) and direction. It is typically represented by a line whose length is equal to the speed of the object, and whose direction indicates the direction of motion. Velocity can be measured in various units, such as meters per second (m/s), kilometers per hour (km/h) or feet per second (ft/s).

Absolute velocity of B:

V_B = (-2cosθ, 2sinθ) m/s

Absolute velocity of C:

V_C = (-2cosθ + 2sinθ, 2sinθ) m/s

Absolute angular velocity of rod BC:

ω_BC = 2rad/s

Absolute acceleration of C:

a_C = (-2sinθ, 2cosθ) m/s^2

Absolute acceleration of B:

a_B = (-2sinθ, 2cosθ) m/s^2

Absolute angular acceleration of rod BC:

α_BC = 0rad/s^2

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The capacitive reactance of a DC series circuit increases when its capacitance decreases and vice-versa.

A DC series circuit can be defined as a type of circuit in which all of its resistive components are connected end to end, so as to form a single path for the flow of current.

This ultimately implies that, the same amount of current flows through a direct current (DC) series circuit.

Mathematically, the capacitive reactance of a DC series circuit is given by this formula:

\(X_C = \frac{1}{2\pi fC}\)

Where:

From the above formula, we can deduce that the capacitive reactance of a DC series circuit is inversely proportional to both frequency and capacitance. Thus, the capacitive reactance of a DC series circuit increases when its capacitance decreases and vice-versa.

In conclusion, a capacitor would influence a simple DC series circuit by blocking the flow of direct current (DC) through it.

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The same as when rounding to three decimal places, we also round a number to three significant figures.

For three digits, we start counting with the first non-zero digit. Next, we round the final digit. Any empty spaces to the right of the decimal point are filled with zeros. As a result, when 1,500 is written without a decimal point, the two trailing zeros are not important; instead, the number has two significant figures. The fact that 1,500.00 has a decimal point, however, makes all six digits important. Significant figures allow us to demonstrate a number's accuracy. A number's integrity is compromised if it is used to express something that is outside of its real range of measurement.

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The deceleration required to comply with the speed limit before being caught by the camera is 6.094 meters per square second.

Let assume that the motor car decelerates at constant rate. Given the initial and final speeds (\(v_{o}\), \(v\)), in meters per second, and travelled distance (\(s\)), in meters, the deceleration (\(a\)), in meters per square second, is determined by this formula:

\(a = \frac{v^{2}-v_{o}^{2}}{2\cdot s}\)

If we know that \(v_{o} = 25\,\frac{m}{s}\), \(v = 40\,\frac{m}{s}\) and \(s = 80\,m\), then the deceleration required to comply with the speed limit is:

\(a = \frac{\left(25\,\frac{m}{s} \right)^{2}-\left(40\,\frac{m}{s} \right)^{2}}{2\cdot (80\,m)}\)

\(a = -6.094\,\frac{m}{s^{2}}\)

The deceleration required to comply with the speed limit before being caught by the camera is 6.094 meters per square second.

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A statement that assigns all elements in the array atomic Weights the value 1.003,independent of the size of the array is atomicWeights(:) = 1.0032.

An array is a grouping of identically typed components that are stored in adjacent memory locations and can each be individually referred to using an index to a special identifier. There is no need to declare five distinct variables when declaring an array of five int values (each with its own identifier). are by default not set. This indicates that none of the components in the array have any specific values assigned to them; rather, their contents are unknown at the time the array is declared.

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Answer:

Most cylinder wear occurs at the top of ring travel.

Answer:

1.2 Newtons upwards

Explanation:

because the friction is opposite of the magnitude.

**All of the above** are required on federally controlled waters for boats less than 39.4 feet. This includes a boating license, a safety certificate, and a life jacket for each person on board.

When operating a boat on federally controlled waters, individuals operating boats less than 39.4 feet in length are required to have a **boating license**. This license ensures that the boat operator has the necessary knowledge and skills to navigate safely. Additionally, a **safety certificate** is mandatory, which typically involves completing a boating safety course. This certificate serves as proof that the boat operator has received proper education in boating safety. Finally, each person on board must have a **life jacket** readily accessible and appropriate for their size. This ensures the safety of everyone in case of an emergency. These requirements are put in place to promote safe boating practices and prevent accidents on federally controlled waters.

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Answer:

The answer is all of the above.

Mechanical skills are highly important to engineering because:

Engineering principles are always applied to machines. Mechanical engineers need to be able to understand how machines work and how to design them.

Engineers will need to be creative to build effective machines. Mechanical engineers need to be able to come up with new and innovative designs that meet the needs of their clients.

Engineers will need problem solving skills. Mechanical engineers need to be able to identify and solve problems that arise during the design and construction of machines.

Engineers must maintain the machines that they create. Mechanical engineers need to be able to troubleshoot and repair machines when they break down.

In addition to mechanical skills, mechanical engineers also need to have strong math and science skills, as well as the ability to work well with others.

Explanation:

Solution :

A wide band pass filter has (Q < 10)

Gain in dB of overall filter = 17 dB

\($F_L = 20 \ Hz $\)

\($F_H = 20 \ kHz$\)

\($17 = 10 \log A_0$\)

\($1.7 = \log A_0 $\)

\($A_0 = 5.47 $\)

\($F_L = 20 \ Hz $\)

\($F_L = \frac{1}{2 \pi R_1C_1}$\)

Let \($C_1 = 0.1 \mu F$\)

\(20 = \frac{1}{2 \pi \times R_1 \times 0.1 \times 10^{-6}}$\)

\($R_1 = \frac{10^6}{4 \pi \times 0.1} $\)

\($R_1 = 79.6 \ k \Omega$\)

\($F_H = 20 \ kHz$\)

\($F_H = \frac{1}{2 \pi R_2C_2}$\)

\($20 \times 10^3 = \frac{1}{2 \pi R_2 \times 0.1 \times 10^{-6}}$\)

\($R_2 = \frac{10^3}{4 \pi \times 10^3} $\)

\($R_2 = 79.6 \ \Omega$\)

Since the overall gain off filter is 5.47

\($A_0 = 5.47$\)

\($A_0 = A_{01} \times A_{02}$\)

\($A_0 = 3.4 \times 1.6$\)

\($A_0 \sim 5.47$\)

\($A_{01} = 1+ \frac{R_F_1}{R_i_1}$\)

\($3.4 = 1+ \frac{R_F_1}{R_i_1}$\)

\($2.4 = \frac{10 k}{R_i_1}$\)

\($R_i_1 = 4166 \ \Omega$\)

\($R_i_1 = 4.1 \ k \Omega $\)

\($A_{02} = 1+ \frac{R_F_2}{R_i_2} $\)

\($1.6 = 1+ \frac{R_F_2}{R_i_2}$\)

\($0.6 = \frac{R_F_2}{R_i_1}$\)

\($R_i_2 = \frac{0.1 k}{0.6}$\)

\($R_i_2 = 166 \ \Omega $\)

Answer:

a) The amount by which this specimen will elongate in the direction of the applied stress is 0.466 mm

b) The change in diameter of the specimen is  - 0.015 mm

Explanation:

Given the data  in the question;

(a) The amount by which this specimen will elongate in the direction of the applied stress.

First we find the area of the cross section of the specimen

A = \(\frac{\pi }{4}\) d²

our given diameter is 20.5 mm so we substitute

A = \(\frac{\pi }{4}\) ( 20.5 mm )²

A = 330.06 mm²

Next, we find the change in length of the specimen using young's modulus formula

E = σ/∈

E = P/A × L/ΔL

ΔL = PL/AE

P is force ( 46300 N), L is length ( 201 mm ), A is area ( 330.06 mm² ) and E is  elastic modulus (60.5 GPa) = 60.5 × 10⁹ N/m² = 60500 N/mm²

so we substitute

ΔL = (46300 N × 201 mm) / ( 330.06 mm² × 60500 N/mm² )

ΔL =  0.466 mm

Therefore, The amount by which this specimen will elongate in the direction of the applied stress is 0.466 mm

(b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

Using the following relation for Poisson ratio

μ = -  Δd/d / ΔL/L

given that Poisson's ratio of the metal is 0.33

so we substitute

0.33 = -  Δd/20.5 / 0.466/201

0.33 = -  Δd201 / 20.5 × 0.466

0.33 = - Δd201  / 9.143

0.33 × 9.143 =  - Δd201

3.01719 = -Δd201

Δd = 3.01719 / - 201

Δd  = - 0.015 mm

Therefore, The change in diameter of the specimen is  - 0.015 mm

Answer:

A

Explanation:

Answer:

a

Explanation:

Answer:

The electric ceiling fan was invented in 1882 by engineer and inventor, Philip Diehl. He had earlier invented an electric sewing machine and adapted the motor from this invention to create the ceiling fan. He called his invention the “Diehl Electric Fan” and it was such a success that he soon had many other people competing with him.

Explanation:

Answer:

13.335 CM (1 ft, 1.335 cm)

I am 80% sure this is the answer, but i am not too keen on math so if i am wrong let me know and i will try my best to fix it!

I hope this helped! Have a good day :]

Answer:

b. American Society of Mechanical Engineers

Explanation:

The "American Society of Mechanical Engineers" (ASME) is an organization that ensures the development of engineering fields. It is an accreditation organization that ensures parties will comply to the ASME Boiler and Pressure Vessel Code or BPVC.

The BPVC is a standard being followed by ASME in order to regulate the different pressure vessels and valves. Such standard prevents boiler explosion incidents.

Both reference counting and mark-and-sweep are widely used garbage collection techniques, and each has its own advantages and limitations.

The choice between the two depends on the specific requirements and constraints of the system. Let's examine the characteristics of both methods:

1. Reference Counting:

- Reference counting works by maintaining a count of the number of references to an object. When the count reaches zero, indicating that no references exist, the object can be safely deallocated.

- The main advantage of reference counting is its efficiency in reclaiming memory. Objects are deallocated immediately when their reference count reaches zero, which helps in minimizing memory usage.

- However, reference counting can be problematic in the presence of circular references, where objects reference each other in a cycle. In such cases, reference counting alone cannot reclaim memory, leading to memory leaks. Additional techniques, like cycle detection algorithms, are required to handle circular references.

2. Mark-and-Sweep:

- Mark-and-sweep is a tracing garbage collection algorithm that identifies and reclaims objects that are no longer reachable by traversing the object graph.

- One advantage of mark-and-sweep is its ability to handle circular references effectively. It can detect and collect objects involved in circular references as long as they are not reachable from the root set.

- Mark-and-sweep introduces a pause during garbage collection as it traverses the object graph, which may cause temporary interruptions in the execution of the program.

- Additionally, mark-and-sweep can lead to fragmentation of the memory space, as it collects objects based on reachability rather than compacting the memory.

Considering these aspects, it is difficult to definitively claim that one method is superior to the other in all scenarios. The choice between reference counting and mark-and-sweep depends on various factors, such as the nature of the application, memory usage patterns, performance requirements, and available resources.

In the context of tombstone management, both reference counting and mark-and-sweep can be viable options. Reference counting can be effective if the system does not involve circular references or if circular references can be handled separately. On the other hand, mark-and-sweep can handle circular references but may introduce some pause time during garbage collection.

Ultimately, the decision should be based on careful analysis of the specific requirements, constraints, and trade-offs of the system at hand. It may even be possible to combine elements of both techniques or utilize alternative garbage collection algorithms tailored to the tombstone management needs of the system.

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Answer:

The Dependent Variable is the Effect, Its value depends on changes from the Independent Variable

Hope this Helps

The non-linearity error as a percentage of the full-scale reading at 100 degree C  is - 0.89%.

Non-linearity error is defined as the biggest deviation from a straight line linking the output signal to the applied force, known as the "best straight," that the calibration curve makes. In order to eliminate second order nonlinearity, the sensor signal conditioning system feeds a portion of the output signal back to the sensor.

If EMF and Celsius temperature have a proportional relationship, then the recorded value of EMF at 100 °C will be understood to be a temperature of.

T/5.268 = 400/21.846

T = 5.268(400/21.846) ≈ 96.457 °C

As a percentage of the full scale, the mistake is

(96.457 -100)/400 × 100% ≈ -0.886%

Thus, the non-linearity error as a percentage of the full-scale reading at 100 degree C  is - 0.89%.

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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.

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A full binary tree with 1000 internal vertices has 2(1000) = 2000 edges.

A full binary tree is a binary tree in which each internal vertex has exactly two children, and each leaf node has no children. In a full binary tree, the number of leaf nodes (L) is one more than the number of internal vertices (I). Therefore, we have:

L = I + 1

Given that there are 1000 internal vertices, we can calculate the number of leaf nodes:

L = 1000 + 1
L = 1001

Now, we can find the total number of vertices (V) in the tree by adding the number of internal vertices and leaf nodes:

V = I + L
V = 1000 + 1001
V = 2001

Since a tree has one less edge than the total number of vertices, the number of edges (E) in the tree is:

E = V - 1
E = 2001 - 1
E = 2000

So, a full binary tree with 1000 internal vertices has 2000 edges.

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