Company A had an engineering job to be given to a subcontracting company. The subcontracting took the job and a formal contract was signed between the two parties. While the project was ongoing, some technical difficulties faced by the subcontracting company forced the project to be stopped for a period of 1 month. Since the project was stalled for 1 month the company A couldn’t complete the project, and couldn’t deliver the project to the client. The client levied a fine on the contracting company. Company A asked for compensation for the delay of work by the subcontracting company wherein in the formal contract there is no mention that the fine can be levied on any delay of work. The two companies had a dispute and company A had refused to conclude the contract. Apply applicable two Bahrain contract laws in this scenario to have a dispute resolution and come up with an appropriate conclusion to the case.

Technician b is correct many map sensors the signal voltage should increase when the vacuum at the sensor decreases.

An area devoid of matter is called a vacuum. The adjective vacuus, which means "empty" or "void," is the source of the word. A region with a gaseous pressure significantly below atmospheric pressure might be thought of as an approximate representation of such a vacuum. vacuum, space where there is no matter present or where the pressure is so low that any particles present have no impact on any processes occurring there. It is measured in units of pressure and is a state that is significantly lower than the average atmospheric pressure. Even though the finest vacuum is in space, scientists have created vacuum chambers and vacuum pumps to imitate this environment. A vacuum chamber is a small enclosure with no air inside of it.

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False. While covering the items may provide some protection, storing parts outside still poses environmental risks.

Exposed parts may become damaged by weather elements such as wind, rain, and extreme temperatures, which can lead to contamination or degradation. Additionally, outdoor storage increases the risk of theft, vandalism, or accidental damage. Storing parts outside also contributes to the accumulation of waste and litter, which can harm wildlife and the environment. Therefore, it is important to consider proper storage methods and environmental regulations to mitigate any potential risks associated with outdoor storage of parts.

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Answer:Size reduction

Cooling

Mixing

Explanation:

Based on the image attached, the reading of micrometer is= 15.20.

You will see a line of numbers running down the barrel of your micrometer, starting with the barrel scale. On the barrel scale, each number corresponds to 0.100. Looking at the barrel, 1 equals 0.100, 2 equals 0.200, 3 equals 300, and so on. The distance between each tick mark and the larger numbers on the barrel is 0.025, or 25 thousandths. 

Note that micrometer is one that is also referred to as a micrometer screw gauge—is a tool with a calibrated screw that is frequently used for precise measurement of components in mechanical engineering, and others.

Looking at the image, you will see the stop ends at 15 and  and 20 so adding them together will be option A. 15.20.

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Answer: sparks and/or electrical currents

Explanation:

Answer:

yes

Explanation:

blueprint of the construction is a prediction of project its is slightly auto cad

Answer:

Technician B

Explanation:

Answer:

liquifid gas

Explanation:

when gas is introduced to a certain chemical it ends up slowly liquifying itself.

Answer:

Food and drug admin

Explanation:

The Vin(t) has an amplitude range of 0 to 5 and 0 to 2.5 for a period of 1 millisecond (ms). The time increments of 10 microseconds (us) must be plotted between the values of 0 to 1ms. Consequently, there are 100,000 data points in 1ms, with 10us intervals between each data point.

Part 1 Recap and Analysis Part 1 was concerned with the following circuit as shown below.

Vin (t) is fed into the high-pass filter, and Vout (t) is produced at the other end. The output voltage of this high-pass filter was obtained and examined in the frequency domain. To begin, the following variables were used:

RC = 1 x 10-4 s, R = 1 x 103 Ω, and C = 1 x 10-7 F.

Then, using the function h(f), the frequency response was defined as follows: H (f) = h (f)/h (0) = (RCf)/(1 + RCf). The magnitude response, H (f), and phase response, (f), were derived from this expression. Using MATLAB, both the phase and magnitude response were plotted against the frequency of the input signal.

The cutoff frequency (fc) was determined to be 1000 Hz, and the bandwidth (B) was calculated to be 1 kHz. The filter is considered a high-pass filter since it has a 1st order response and is capable of passing signals at frequencies above its cutoff frequency while blocking signals below that frequency. The low frequencies and high frequencies are referred to as noise and signal, respectively.

Vin(t) Graphical RepresentationThe first step is to plot the function Vin(t) mathematically. Vout(t) is defined by the transfer function H(f), which is derived from Vin(t).

The first step is to plot Vin(t), which is given by:

Vin(t) = 5.0sin(wt + 0) + 1.0sin(wt + 0) + 2.5sin(w3t + 0) On the MATLAB Command Window, enter the following code: t = 0:0.00001:0.001; Vin = 5*sin(8000*pi*t)+ 1*sin(29000*pi*t)+ 2.5*sin(242000*pi*t); plot(t,Vin) xlabel('time (s)') ylabel('Amplitude (V)') title('Vin(t) Plot')

Output: The resultant Vin(t) is graphed below.

The initial part oscillates between 0 and 5, and the last section between 0 and 2.5. In other words, the function Vin(t) is made up of three components with different amplitudes and frequencies.

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Answer:

10 ft

Explanation:

The maximum rating of the motor branch-circuit short-circuit and ground-fault protective device for a 7 1/2-horsepower, 208-volt, 3-phase squirrel-cage induction motor using time-delay fuses is approximately 45 A.

To determine the maximum rating of the motor branch-circuit short-circuit and ground-fault protective device for a 7 1/2-horsepower, 208-volt, 3-phase squirrel-cage induction motor, we can use the National Electrical Code (NEC) guidelines.

Step 1: Calculate Full Load Current (FLA)

The full load current (FLA) of a motor can be calculated using the formula:

\(\mathrm{FLA = \frac{Horsepower \times 746}{\sqrt{3}\times voltages \times efficiency \times power factor } }\)

Since efficiency and power factor are not given, we will assume typical values of 0.85 for efficiency and 0.85 for power factor.

Given:

Horsepower (HP) = 7.5 HP

Voltage = 208 V

Efficiency = 0.85

Power Factor = 0.85

Substitute these values into the formula:

\(\mathrm{FLA = \frac{7.5 \times 746}{\sqrt{3}\times 208 \times 0.85 \times 0.85 } }\)

Step 2: Determine the Maximum Rating:-

According to NEC guidelines, the maximum rating of the protective device can be calculated by multiplying the FLA by a percentage that corresponds to the type of motor and the type of protective device.

For a squirrel-cage induction motor with time-delay fuses, the percentage is typically around 175%.

Maximum Rating = FLA × 175%

Maximum Rating = 25.76A × 1.75 ≈ 45.12A

Rounded to the nearest standard fuse size, the maximum rating of the protective device is 45 A.

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The mole ratio of the distillate to the bottoms is 16.24. Distillation is the process of separation of components in a mixture based on their different boiling points.

The feed consisting of 74% ethane and 26% octane is subjected to a distillation unit where the bottoms contain 95% of the heavier component and the distillate contains 99% of the lighter component. All percentages are in moles.To determine the mole ratio of the distillate to the bottoms, we need to calculate the number of moles of ethane and octane in the feed, distillate, and bottoms. Let's consider 100 moles of the feed.The feed contains 74 moles of ethane and 26 moles of octane. The distillate contains 99 moles of ethane, and the bottoms contain 5% of ethane. So the bottoms contain 69.5 moles of octane. Therefore, the mole ratio of the distillate to the bottoms = moles of ethane in the distillate / moles of octane in the bottoms= 99 / 69.5 = 1.424 rounded to two decimal places= 1.42.The mole ratio of the distillate to the bottoms is 1.42 or 16.24.

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The maximum velocity (u_max) in the parabolic laminar flow is 12 cm/s.

In the problem statement, it is given that the incompressible steady flow is uniform with u = U0 = 8 cm/s in the inlet.

Downstream, the flow develops into a parabolic laminar profile with u = az(z0 - z). The fluid is SAE 30 oil at 20°C, and z0 = 4 cm.

First, we need to find the dynamic viscosity of SAE 30 oil at 20°C. SAE 30 oil has a kinematic viscosity (ν) of approximately 300 cSt (centistokes) at 20°C.

To convert this to dynamic viscosity (μ), we need to multiply by the density (ρ) of the oil:

μ = ν * ρ

The density of SAE 30 oil is approximately 0.89 g/cm³ (890 kg/m³). Since 1 cSt is equal to 1 × 10⁻⁶ m²/s, the kinematic viscosity in SI units is 300 × 10⁻⁶ m²/s.

Now, let's convert the density to SI units:

ρ = 890 kg/m³ = 0.89 g/cm³

Thus, the dynamic viscosity (μ) can be calculated as follows:

μ = (300 × 10⁻⁶ m²/s) * (890 kg/m³) = 0.267 kg/(m*s)

Now, we need to find the maximum velocity (u_max) in the parabolic laminar flow, which occurs at the center of the plates (z = z0/2):

u_max = a * z0/2 * (z0 - z0/2)

Since the flow is incompressible, the mass flow rate (Q) remains constant throughout. We can equate the mass flow rate at the uniform flow (Q_inlet) with the mass flow rate at the parabolic flow (Q_parabolic):

Q_inlet = Q_parabolic

ρ * U0 * A_inlet = ∫[ρ * a * z * (z0 - z) * A_parabolic] dz

The area A_inlet and A_parabolic both can be represented as A = b * z, where b is the width of the parallel plates, and z is the distance between the plates.

Therefore, the equation simplifies to:

U0 * b * z0 = ∫[a * z * (z0 - z) * b] dz, with integration limits 0 to z0

U0 * z0 = ∫[a * z * (z0 - z)] dz, with integration limits 0 to z0

8 cm/s * 4 cm = a * ∫[z * (4 cm - z)] dz, with integration limits 0 to 4 cm

32 cm²/s = a * ∫[4z - z²] dz, with integration limits 0 to 4 cm

Now we can integrate and apply the limits:

32 cm²/s = a * [2z² - (1/3)z³] | (0 to 4 cm)

32 cm²/s = a * [(2 * 4² - (1/3) * 4³) - 0]

32 cm²/s = a * (32 - 64/3)

32 cm²/s = a * (32 - 21.33)

32 cm²/s = a * 10.67 cm²

Now we can solve for 'a':

a = 32 cm²/s / 10.67 cm² = 3 cm/s

Finally, we can find the maximum velocity (u_max) at the center of the plates

Now that we have the value of 'a' (3 cm/s), we can find the maximum velocity (u_max) at the center of the plates (z = z0/2):

u_max = a * z0/2 * (z0 - z0/2)

u_max = 3 cm/s * (4 cm)/2 * (4 cm - 4 cm/2)

u_max = 3 cm/s * 2 cm * 2 cm

u_max = 12 cm/s

Thus, the maximum velocity (u_max) in the parabolic laminar flow is 12 cm/s.

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Answer:

The answer is 960 kg

Explanation:

Solution

Given that:

Assume the initial dye concentration as A₀

We write the expression for the dye concentration for one hour as follows:

ln (C₁) = ln (A₀) -kt

Here

C₁ = is the concentration at 1 hour

t =time

Now

Substitute 480 g for C₁ and 1 hour for t

ln (480) = ln (A₀) -k(1) ------- (1)

6.173786 =  ln (A₀) -k

Now

We write the expression for the dye concentration for three hours as follows:

ln (C₃) = ln (A₀) -k

Here

C₃ = is the concentration at 3 hour

t =time

Thus

Substitute 480 g for C₃ and 3 hour for t

ln (120) = ln (A₀) -k(3) ------- (2)

4.787492 = ln (A₀) -3k

Solve for the equation 1 and 2

k =0.693

Now

Calculate the amount of blue present initially using the expression:

Substitute 0.693 for k in equation (2)

4.787492 = ln (A₀) -3 (0.693)

ln (A₀) =6.866492

A₀ =e^6.866492

= 960 kg

Therefore, the amount of the blue dye present from the beginning is  960 kg

Volume of the system (N2) at all states;The first state; \(T1 = 100 °C, P1 = 600 kPa, m = 2.4 kg\) From the ideal gas law;\(PV = mRT;V1 = (mRT1) / P1;V1 = (2.4 kg) (0.2968 kPa m³ / kg K) (100 + 273.15) K / (600 kPa)V1 = 0.2533 m³\) The second state;

The constant-pressure process;T2 = 300 °C, P2 = 600 kPa From the ideal gas law;\(V2 = (mRT2) / P2;V2 = (2.4 kg) (0.2968 kPa m³ / kg K) (300 + 273.15) K / (600 kPa)V2 = 0.5066 m³\)The third state; The isothermal process;\(T3 = 300 °C, P3 = 400 kPa\) From the ideal gas law;\(V3 = nRT / P3;V3 = (2.4 kg) (0.2968 kPa m³ / kg K) (300 + 273.15) K / (400 kPa)V3 = 0.6332 m³\)The fourth state; The polytropic process;

P4 = 300 kPa, n = 1.4 From the polytropic process equation\(;P1V1ⁿ = P2V2ⁿ;V4 = V3 (P3 / P4)^(1/n);\)V4 = 0.6332 m³ (400 kPa / 300 kPa)^(1/1.4)V4 = 0.8088 m³b) Final temperature;The final temperature is the same as the initial temperature (isothermal process). \(V4 = 0.6332 m³ (400 kPa / 300 kPa)^(1/1.4)V4 = 0.8088 m³b)\)T4 = T3 = 300 °Cc) Boundary work for each process;The boundary work is given by;\(W = ∫PdVFor the first process, the boundary work is;W1 = ∫PdV;W1 = P1(V2 - V1)W1 = (600 kPa) ((0.5066 m³) - (0.2533 m³))W1 = 152.997 kJ\)For the second process, the boundary work is;

\(W2 = ∫PdV;W2 = P2(V3 - V2)W2 = (600 kPa) ((0.6332 m³) - (0.5066 m³))W2 = 45.986 kJ\)For the third process, the boundary work is;

\(W3 = ∫PdV\)\(;W3 = RT ln(P3 / P4)W3 = (2.4 kg) (0.2968 kPa m³ / kg K) ln(400 kPa / 300 kPa)W3 = 23.125 kJd)\) Net work (sum of all individual work components) for the entire thermodynamic process;The net work is the sum of all individual works;\(W_net = W1 + W2 + W3W_net = 152.997 kJ + 45.986 kJ + 23.125 kJW_net = 222.108 kJe)\)Show the processes on a P vs V diagram;The processes are shown on the diagram below;

The first process (1 to 2) is a horizontal line at a pressure of 600 kPa.The second process (2 to 3) is a vertical line at a volume of 0.5066 m³.The third process (3 to 4) is a curve with a polytropic exponent, n = 1.4.

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The analog computer circuit can be designed using integrator and differentiator circuits, combined with an op-amp summer circuit, to represent the derivatives and solve the equation in an analog manner.

To design an analog computer circuit that solves the given differential equation, we can use operational amplifiers (op-amps) and passive components. The circuit can be divided into two parts: an integrator and a differentiator.

The integrator circuit, using an op-amp and capacitors, integrates the input signal twice, representing d²v_o/dt² and dv_o/dt. The differentiator circuit, using resistors and capacitors, differentiates the output of the integrator, representing 2dv_o/dt.

The output of the integrator, differentiator, and a feedback resistor are combined in an op-amp summer circuit to generate the final output v_o. The input signal 10sin(4t) is applied to the circuit.

By setting appropriate initial conditions (v_o(0) = 0 and v'_o(0) = 0), the circuit can solve the given differential equation in an analog manner.

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Answer:

Weight Percentage of Ash = 3.4

Explanation:

Given - Coal containing 21% ash is completely combusted, and the ash is 100% removed in a water contact scrubber. If 10,000 kg of coal is burned per hour with a scrubber flow rate of 1.0 m3/min.

To find - the weight percentage of the ash in the water/ash stream leaving the scrubber is most nearly ?

Solution -

Given that,

Coal Burned Rate = 10,000 kg/hr

                              = \(\frac{10,000}{60 min} * 1 hr *\frac{kg}{hr}\)

                              = 166.6666 kg/min

⇒Coal Burned Rate = 166.6666 kg/min

Now,

Given that,

Ash content in coal = 21 %

⇒Ash in (coal that burned) = 166.6666 × \(\frac{21}{100}\) kg/min

                                             = 34.9999 ≈ 35 kg/min

⇒Ash in (coal that burned) = 35 kg/min

Now,

We know,

Density of water = 1000 kg/m³

Now,

Water flow Rate = \(1\frac{m^{3} }{min} * density\)

                           = 1000 kg/min

⇒Water flow Rate = 1000 kg/min

Now,

Total Mass flow Rate of (Water + Ash stream) = ( 1000 + 35) kg/min

                                                                           = 1035 kg/min

⇒Total Mass flow Rate of (Water + Ash stream) = 1035 kg/min

So,

Weight Percentage of Ash = (Weight of Ash ÷ Total weight of Stream) × 100

                                            = (35 ÷ 1035) × 100

                                            = 3.38 ≈ 3.4

∴ we get

Weight Percentage of Ash = 3.4

Answer:

a) When the switch is off and there's no battery supply, the electric potential difference, or electromotive force E(t), is zero. That means the right-hand side of the given differential equation is zero. Therefore, the solution to the homogeneous differential equation represents the charge on the capacitor:

L d²Q/dt² + R dQ/dt + 1/C Q = 0

In this case, since there's no initial charge or current supplied, Q(t) = 0 for all t.

b) When the switch is turned on and a 12-Volt battery is connected:

i. The method of Undetermined Coefficients:

We can solve this by proposing a particular solution that has the same form as the non-homogeneous term, E(t). As E(t) = 12 volts is a constant, we propose Q(t) = A as a constant.

After substituting Q(t) = A into the equation, we would be able to find the value of A, which would give us the particular solution. The general solution would then be the sum of this particular solution and the solution to the homogeneous equation (obtained from part (a)).

ii. The method of Variation of Parameters:

In this method, we would make use of the solutions of the homogeneous differential equation. After finding these, we would propose a solution for the non-homogeneous differential equation in terms of these solutions, and a pair of functions (u and v) to be determined. We then substitute this proposal into the differential equation to obtain a set of two new first-order differential equations for u and v.

c) Once we've found the charge Q(t) in part (b), we can find the current I(t) by differentiating Q(t), as I(t) = dQ/dt.

d) With the given initial conditions (Q = 0.001 C, I = 0 A), we can substitute these into the general solution and its derivative obtained in part (b). We would then solve the resulting system of two equations to find the constants involved, allowing us to determine the specific solution for these initial conditions.

Explanation:

Complex question. Answer depends on data provided and format of equations provided.

Technician A says a defective PCV valve may cause rough idle operation  is correct and so is  Tech B: So C)both A and B are correct.

If a PCV valve is known to be stuck open, then the PCV valve is one that can be said to be faulty.

Note that a faulty PCV valve can lead to more air going into the combustion chamber that can leads to lean air-fuel ratio.

Therefore, Technician A says a defective PCV valve may cause rough idle operation  is correct and so is  Tech B: So C)both A and B are correct.

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Technician A says a defective PCV valve may cause rough idle operation. Technician B says the PCV system depends upon a properly sealed engine to perform correctly. Who is correct?

A)A only

B)B only

C)both A and B

D)neither A nor B

Answer:

6.564 is the answers i learned that in the american school i live there

False. Fire extinguishing equipment cannot be frozen according to Ref. 213/91.

According to Ref. 213/91, fire extinguishing equipment cannot be frozen. Fire extinguishers are essential safety devices designed to combat fires effectively. They contain pressurized agents that are specifically formulated to extinguish different types of fires. Freezing temperatures can significantly impair the functionality of fire extinguishers and render them ineffective in emergency situations.

When fire extinguishing equipment freezes, several issues can arise. First, the contents of the extinguisher may expand as they freeze, potentially leading to ruptures or leaks in the container. This can cause the extinguisher to malfunction or become hazardous when used. Second, freezing temperatures can affect the performance of the extinguishing agent itself. Certain agents, such as water-based solutions, can solidify or lose their effectiveness when exposed to extreme cold.

It is crucial to store fire extinguishers in suitable environments that are above freezing temperatures. This ensures that the equipment remains in optimal condition and is ready for immediate use during emergencies. Regular inspections and maintenance are also essential to identify any signs of damage or deterioration that may compromise the functionality of fire extinguishers.

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Explanation:

\( \frac{1}{8} + \frac{1}{2} \\ 1.6 + 1.4 = 3 \\ \frac{1}{3} + \frac{1}{9} \\ 2.25 + 2 = 4.25 \: ohm\)

R total = 4.25 ohm

I total = Vt/Rt

I total= 17/4.25= 4 A

Ix= 600 mA

\( \frac{9}{9 + 3} \times 4 = 3\\ \frac{2}{2 + 8} \times 3 = 0.6a \\ = 0.6 \: milli \: amper\)

Answer:

Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.

Note: The diagram referred to in this question is attached as a file below.

Answer:

The block descended a distance of 9.75m from the instant the brake is applied until it stops.

Explanation:

For clarity and easiness of expression, the calculations and the Free Body Diagram are contained in the attached file. Check the attached file below.

The block descended a distance of 9.75 m

What the question wants is to write a code that would find and display the average of two numbers given by a user.

The code or program has been written and has been given below in C + +

#include <iostream>

using namespace std;

int main() {

   int a, b, sum;

   double avg;

   cout << "Enter the first integer number: ";

   cin >> a;

   cout << "Enter the second integer number: ";

   cin >> b;

   sum = a + b;

   avg = (double)sum / 2;

   cout << "\nThe average of your numbers is: " << avg << "\n";

   return 0;

}

Enter the first integer number: 2

Enter the second integer number: 4

The average of your numbers is: 3

A simple explanation of the code is that it makes use of mathematical operators to perform the calculations by first adding up the numbers and then finding the average of the two numbers by dividing the sum of the numbers with two which is the number of elements.

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Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= \(density\times g\times \frac{h}{2}\)

By putting values, we get

= \(1000\times 9.8\times \frac{12.2}{2}\)

= \(1000\times 9.8\times 6.1\)

= \(59780\)

hence,

The average force will be:

= \(Pressure\times Area\)

= \(59780\times 3.6\times 12.2\)

= \(2625537 \ N\)

Or,

= \(2625 \ kN\)

Answer:

B....................

Answer:

AC

Explanation:

One situation when alternating current would work better than direct current is if the operator is encountering magnetic arc blow.

Current works best when the operator  encounters magnetic arc blow is AC

Magnetic arc blow is simply defined as the arc deflection due to the warping of the magnetic field that is produced by electric arc current.

This is caused as a result of the following;

- if the material being welded has residual magnetism at an intolerable level

- When the weld root is being made, and the welding current is direct current which indicates constant direction and maintains constant polarity (either positive or negative).

Thus, Current works best when the operator  encounters magnetic arc blow is AC

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