Are the stack temperature and oxygen reasonable for these operating conditions? if not, what oxygen and stack temperature would you recommend if not? what is the efficiency with your chosen values?

Complete Question

In Problems 69 through 72 you are given the equation(s) used to solve a problem. For each of these,

(a) Write a realistic problem for which this is the correct equation(s).

The electric field due to a point charge at distance of 1 um is given as 1.5x10^6 N/C

E = (9*10^9 Nm^2/C^2*N*1.6*10^-19 C)/(1*10^-6)^2

E = 1.5x10^6 N/C

Hence, Electric field is 1.5x10^6 N/C

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The left hand plays a steady one-two beat in duple meter, while the right hand emphasizes notes that fall in between these beats. The name for this technique is syncopation.

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At F the object moving to the left, if the positive x-direction corresponds to its rightward movement.

Δx = [tex]X _{2}[/tex] - [tex]X_{1}[/tex]

where,

Δx = Change in position

[tex]X _{2}[/tex] = Final position

[tex]X_{1}[/tex] = Initial position

If Δx is positive, the object is moving forwards ( Right ).

If Δx is negative, the object is moving backwards ( Left ).

If the curve is moving from top to bottom Δx will be negative and if the curve is moving from bottom to top Δx will be positive.

So, in this given curve, at the points B, C and D, the curve is moving from bottom to top meaning that the object is moving to the right and at the point F, the curve is moving from top to bottom meaning that the object is moving to the left.

Therefore, At F the object moving to the left, if the positive x-direction corresponds to its rightward movement.

The given question contains no distance - time graph. I considered the attached image as the graph.

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After performing gel electrophoresis on the pcr product, you see that one sample has a 200 bp band and the other does not this shows that 200 bp band is significantly larger than the size of primer dimers unless the primers are 100 bp long, they are not visible.

Gel electrophoresis is the process used to evaluate the success of the PCR reaction.A closer examination of the suitability of agarose gel electrophoresis is done for assessing PCR product yield.

PCR is a series of chemical reactions that take place outside a living cell

Use of 2% agarose gel and 100bp marker and spreading the marker strips properly with 80-90V is done to avoid heating.

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The density of a proton is 2.87 x 10^19 kg/m^3.

The density of a proton can be calculated by dividing its mass by its volume i.e.,

Density = Mass/Volume

The mass of a proton is given to be 1.67 x 10^-27 kg.

As the proton is assumed to be a sphere, its volume will be 4/3 π r^3.

Thus, density = 1.67 x 10^-27/ (4/3 x π x r^3)

Putting radius of proton(r) = 2.4 fm = 2.4 x 10^-15 m

The density comes out to be 2.87 x 10^19 kg/m^3.

Thus, the density of a proton is 2.87 x 10^19 kg/m^3.

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Magnetoencephalography (MEG) allows researchers to examine the brain's magnetic field. Which helped researchers to understand how certain tasks influence brain activity.

It isn't. Invaluable tool to study the large scale brain activity. And interactions with the body.

MEG was developed in 1960s .

In this technique weak magnetic forces associated with electrical activities of brain are monitored .

It collects activity from neurons visualized as brain waves and it is used with MRI in magnetic source imaging.

Med fields passes through the head without any distortion.

It provides a Spatial and temporal resolution.

Magnetoencephalography (MEG) allows researchers to examine the brain's magnetic field.

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The current that can flow in the secondary is 23 A

According to Transformation ratio,

[tex]V_{p}[/tex] / [tex]V_{s}[/tex] = [tex]I_{s}[/tex] / [tex]I_{p}[/tex]

where,

[tex]V_{p}[/tex] = Voltage across the primary coil

[tex]V_{s}[/tex] = Voltage across the secondary coil

[tex]I_{s}[/tex] = Current in the primary coil

[tex]I_{p}[/tex] = Current in the secondary coil

Given that,

[tex]V_{p}[/tex] = 110 V

[tex]V_{s}[/tex] = 24 V

[tex]I_{p}[/tex] = 5 A

[tex]V_{p}[/tex] / [tex]V_{s}[/tex] = [tex]I_{s}[/tex] / [tex]I_{p}[/tex]

110 / 24 = [tex]I_{s}[/tex] / 5

[tex]I_{s}[/tex] = 22.9 = 23 A

Transformation ratio is the ratio of secondary voltage to primary voltage.

Therefore,  The current that can flow in the secondary is 23 A

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The correct answer is an important measure for controlling bias.

An effective way to prevent under-coverage in a memory experiment would be to randomly assign students to two separate groups.

The scientific method that must be repeatable and the data must be empirical for science to be trusted.

When scientists publish their work, they include enough information so that other scientists can repeat the experiment and obtain the same outcomes.

Additionally, in contrast to theories that are not supported by facts, the scientific method is founded on empirical evidence, which is derived from personal observations of the outside world.

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The astronaut's weight on the surface of Mars is 259 N

To solve this exercise, the formula and procedure to be applied is:

W = m * g

Where:

Information about the problem:

Applying the weight formula, we get:

W = m * g

W = 70 kg * 3.7 m/s2

W = 259 N

In physics, gravity is the force of attraction that the earth exerts on all bodies possessing mass by pulling them toward its center.

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The buoyancy force on the gold ball will be more than that of aluminum.

An upward force applied by a fluid against the weight of an item that is partially or totally submerged is known as upthrust, also known as buoyancy. Pressure in a fluid column increases with depth due to the weight of the fluid on top.

Given that two solid metal balls all have the same radius. one is made of aluminum, the other of gold. And the balls are resting at the bottom of a pool of water.

The buoyancy force at the bottom force will be calculated by finding the density of the balls.

Buoyancy force = Volume inside water x Density X g

Density of gold = 19.3 g / cc

Density of aluminium = 2.7 g / cc

Since the density of the gold ball is more than the density of the aluminum ball.

Therefore, the buoyancy force on the gold ball will be more than that of aluminum.

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The number of stars range between 6 * 10^9 stars and 1.2 * 10^10 stars.

We know that a galaxy is composed of many stars. The stars in the galaxy as we saw from the question, using the milky way as a template may have between 200 and 400 billion stars.

The number of stars could be obtained as the product of the number of galaxies and the number of stars in the lower and the upper limits.

Number of stars in the lower estimate = 200 * 10^6 * 30 = 6 * 10^9 stars

Number of stars in the higher estimate = 400  * 10^6 * 30 = 1.2 * 10^10 stars

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A horizontally applied force of 2400 N will accelerate a 400 kg crate at 1 m/s^2.

In the given diagram, the free body diagram (FBD) of a block is shown.

Here the two vertical forces, normal reaction(N) and weight(W) cancel each other while the horizontal forces do not cancel each other and provide acceleration to the block.

According to the given situation,

Applied Force - Frictional Force = ma

Here the frictional force is half the weight of the body 2000 N.

Putting the given values in the above equation,

F - F1 = 400 x 1

F - 2000 = 400

F = 2400 N

Hence, a horizontally applied force of 2400 N will accelerate a 400 kg crate at 1 m/s^2.

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The following statements have italicized word, and has 2 correct  statement and 1 incorrect statement.

a. Taylor's weight is 65 kilograms. : Its correct.

b. I went on a diet and my mass decreased. : Its correct.

c. I went to the Moon and I lost mass, even though I didn't lose any of my matter. : Its incorrect.

The weight is a term shows the mass of the object and the gravity working on it. Example is that a person who feel heavy in earth that same person feel very light on moon cause the gravity will work on that object verry small in comparison to earth.

a. Taylor's weight is 65 kilograms. It is a complete sentence. So its a correct statement.

b. I went on a diet and my mass decreased. : Its correct. Its also a complete sentence. So its a correct statement.

But, c. I went to the Moon and I lost mass, even though I didn't lose any of my matter. This sentence is incorrect because the mass of the person remains same in the moon but the gravity become lesser in the moon so he felt lighter.

So, according to the theory the statement is incorrect.

The correct statement would be:

c. I went to the Moon and I lost weight, even though I didn't lose any of my matter.

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The rectangular resistors with leads specifically made to fit into printed circuit boards are surface mount resistors.

The resistors are the device which resists the motion of current or electricity.

They are electronic devices used in electrical circuits.

The general type of resistors which are rectangular with leads so taht they fit the circuit boards are called surface mount resistors.

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The distance of a point to the yz-plane is given by the absolute value of the x coordinate. So, for point A, the distance is equal to,

Dyz = | -4 | = 4 units

Next, for point B, the distance is equal to,

Dyz = | 3 | = 3 units

Lastly, for point C, the distance is equal to,

Dyz = | 2 | = 2 units

Since point C has the smallest distance to the yz -plane, this point is closest to the yz-plane.

Unlike the regular 2-dimensional Cartesian coordinate system, a three-dimensional coordinate system contains three axes. To plot points in this coordinate system, the point's ordered triple must be known, (x,y,z,).Each coordinate on the ordered triple represents a point that lies on one of the axes

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By using uniform motion, the distance traveled by the car is 216 m.

We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule

vt = vo + a . t

vt² = vo² + 2a . s

s = vo . t + 1/2 . a . t²

where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.

From the question above, we know that

vo = 15 m/s

vt = 21 m/s

t = 12 s

Find the acceleration

vt = vo + a . t

21 = 15 + a . 12

12a = 6

a = 0.5 m/s²

By using the third equation, we can calculate the distance

s = vo . t + 1/2 . a . t²

s = 15 . 12 + 1/2 . 0.5 . 12²

s = 180 + 36

s = 216 m

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By using uniform motion, the object will travel 2.22 meters until it stops.

We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule

vt = vo + a . t

vt² = vo² + 2a . s

s = vo . t + 1/2 . a . t²

where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.

From the question above, we know that

vo = 1.70 m/s

F = -52 N

m = 80 kg

vt = 0 m/s

Find the deceleration

F = m . a

-52 = 80 . a

a = -0.65 m/s²

Find the distance using the second equation

vt² = vo² + 2a . s

0² = 1.7² + 2(-0.65) . s

1.3s = 2.89

s = 2.22 meters

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Sound travels faster in the air if the air temperature is warm

Everything comes down to the chemical composition of air and sound. Sound molecules travel faster through the air particles in warm air. So, at an air temp of 70°F at sea level, the scientific estimation of the sound speed is 767 mph. Sound, on the other hand, travels the fastest in warmer temperatures air. It moves at 740 miles per hour in subzero temperatures.

Although sound travels slower in cooler air temperatures, it can travel further. The waves are refracted from wallets of warm air and returned to the ground as a result. This sound wave then oscillates between cold and warm air and the surface, causing it to travel further.

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According to the given statement is  At address 0x8004, the default data word is 0xA7908CEE.

Little Endian method with a word size of 4 Bytes is the default on ARM processors.

Data is stored in the memory bytes at a time, and access is done in words.

The data must be stored in 4 cells because word size is equal to 4 bytes.

i.e., the data word is retrieved in parallel from 4 cells.

Little endian states that lower addresses have lower bytes and higher addresses have higher bytes.

At location 0x8004, the default data word is 0xA7908CEE.

The sequence in which data is stored in memory is referred to as the "little-endian" convention. According to this convention, the least significant bit (or "littlest" end) is placed at address 0 first, followed by the storage of following bits one at a time.

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The correct answer is  (E)  It could have any value between 2.0 km and 14.0 km depending on how the vectors are oriented.

The magnitude of the resultant displacement could have any value between 2.0 km and 14.0 km depending on how the vectors are oriented.

In physics, displacement (which is a vector quantity) refers to a change in an object's position. When you compute displacement, you compare an object's original and final locations to determine how "out of place" it is. The formula you employ to calculate displacement will be based on the variables presented to you in a particular problem.

The resultant is the vector sum of each individual vector. The different vectors are combined to produce the outcome. By combining the separate displacements using vector addition techniques, the outcome can be calculated.

Therefore the resultant displacement can have any value from 2-14. It depends upon the orientation of displacement vectors.

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The complete question is:

"A physics student adds two displacement vectors with magnitudes of 8.0 km and 6.0 km. Which one of the following statements is true concerning the magnitude of the resultant displacement? A) It must be 10.0 km. B) It must be 14.0 km. C) It could be equal to zero kilometers, depending on how the vectors are oriented. D) No conclusion can be reached without knowing the directions of the vectors. E) It could have any value between 2.0 km and 14.0 km depending on how the vectors are oriented."

From the sun to Alpha Centauri than from the sun to Jupiter it is 105 times further. The age of Alpha Centauri is 5.3 billion years older. It also moves in a totally different way. Even though it is close by, it is only passing.

Alpha Centauri A is around 100,000 miles farther away, 10% more massive, and a little bigger than our Sun. All of the planets' orbits will become more elliptical as a result of the increased mass. The reality that Alpha Centauri A is 50% brighter than our star, placing Earth slightly inside the habitable zone, is much worse for the Earth. Most of Earth's surface is currently uninhabitable due to its average surface temperature of 43 degrees Celsius, according to Universe Sandbox 2. It has been clearly stated to us, though, that a temperature so high would most likely lead to an uncontrollable greenhouse effect by increasing the amount of water vapor in the atmosphere.

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According to the statement:

a) speed is equal to 118 km/h in kilometres per hour.

b) yes it is exceeding 90km/h.

The pace at which an object's position changes in any direction is referred to as speed. Speed is defined as the ratio of something like the distance traversed to the time required to cover that distance. . . Speed is a scalar quantity since it just has a direction and no magnitude..

Speed is determined by dividing the distance by the time. Calculating the units for speed requires knowledge of the units for both time and distance.. The units in this example will be meters per second (m/s), since the distance is measured in meters (m) and the time is measured in seconds (s).

(a)s₂= 118.km/h is its speed in kilometres per hour.

1h = 3600s

1km = 1000m

s₂= (33m/s) *[tex]\left(\frac{60 \times 60}{1000}\right)[/tex]

= 118.8km/h

s₂= 118.km/h

(b) yes it is exceeding 90km/h.

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The bag's velocity when a bag is dropped from a hovering helicopter is 15.68m/s

Its initial velocity is zero. In a free-falling body, the value of the final velocity is obtained through the equation

                                           Vf =u + gt

u = o ( because the bag is dropped, so it falls starting from rest)

Vf = gt

u is initial velocity

Vf is final velocity

t is time

where g is the acceleration due to gravity, 9.8 m/s².

using above values we get

 Vf = (9.8 m/s²)(1.6s)

= 15.68m/s

The bag's velocity when a bag is dropped from a hovering helicopter is 15.68m/s

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By the half-life of radioactive, the remaining mass of radioactive elements is 2.47 pounds.

We need to know about the half-life of the radioactive elements to solve this problem. The radioactive element will decay over time and follow the equation

N = No(1/2)^(t/t'')

where N is the final quantity, No is the initial quantity, λ is the decaying constant, t is time and t'' is the half-life of a radioactive element.

From the question above, we know that

t'' = 9 years

No = 6 pounds

t = 11.5 years

Calculate the remaining mass

N = No(1/2)^(t/t'')

N = 6(1/2)^(11.5/9)

N = 2.47 pounds

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Answer: 600 km west

Explanation:

When looking at total distance/displacement, simply add the two given quantities together. Note- in the future, make sure to pay attention to units and directions of components. For example, if it was 100 km east and then 500 km west, your total would be -100 km west + 500 km west, giving you 400 km west.

The correct answer is 369.2765m

Speed of tortoise = v_t= 16.3cm/s=0.163m/s

Speed of hare= v_h=220.2cm/s=2.202m/s

Rest time of hare is t_r=1.6min

Tortoise wins by distance of x = 108cm=1.08m

Consider that S is the length of a racetrack.

At the conclusion of the race, calculate the tortoise's distance travelled.

S=v_t×t

S=0.163m/s×t ______(1)

Here, t is the race's actual time.

Find out how long the hare spent racing.

t_h=t-t_r

t_h=t-1.6min×(60s/1min)

t_h=(t-2160)s

At the conclusion of the race, determine the hare's distance travelled.

S-x=v_h×t_h

S-1.08m=2.2×(t-2160)s

S=2.2×(t-2160)s+1.08m_____(2)

To determine the race time, substitute equation (1) for equation (2).

0.163m/s×t=S=2.2×(t-2160)s+1.08m

0.163t=2.26t-4752+1.08

4752-1.08=2.26t-0.163t

4750.92=2.097t

4750.92/2.097=t

2265.5=t

To determine the length of the race or the track, substitute the amount of time in equation (1).

S=0.163m/s×t

S=0.163×2265.5

S=369.2765m

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The displacement and velocity values at time, t = 8 seconds are; 11.76 and -0.545 respectively

Part A

The possible values in the table are;

Displacement;

t = 7 seconds, y = 10.02

t = 8 seconds, y = 11.01

t = 9 seconds, y = 13.5

Velocity;

t = 7 seconds, y = 0.06

t = 8 seconds, y = -0.05

t = 9 seconds, y = -1.15

The equation for interpretation which is a form of linear equation is presented as follows;

[tex]y - y_0 = \frac{y_1 - y_0}{x_1 -x_0} \times (x - x_0)[/tex]

Interpolating for the displacement at t = 8.0 seconds is as follows;

[tex]y - 10.02 = \frac{13.5 - 10.02}{9 - 7} \times (8 - 7)[/tex]

Which gives the displacement at t = 8.0 seconds as 11.76

Interpolating for the velocity at t = 8.0 seconds is as follows;

[tex]y - 0.06 = \frac{(-1.15) - 0.06}{9 - 7} \times (8 - 7)[/tex]

Which gives the velocity at t = 8.0 seconds as -0.545

The values for the displacement and velocity obtained by interpolation (11.76 and -0.545) are different from the displacement and velocity values in the table at t = 8 which are; 11.01 and -0.05 respectively

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