air enters the compressor of a simple gas turbine power plant at 708f, 1 atm, is compressed adiabatically to 40 lbf/in.2, and then enters the combustion chamber where it burns completely with propane gas (c3h8) entering at 778f, 40 lbf/in.2 and a molar flow rate of 1.7 lbmol/h. the combustion products at 13408f, 40 lbf/in.2 enter the turbine and expand adiabatically to a pressure of 1 atm. the isentropic compressor efficiency is 83.3% and the isentropic turbine efficiency is 90%. determine at steady state(a) the percent of theoretical air required.(b) the net power developed, in horsepower.

Answer:

The trebuchet was invented in France and was first reported to be used in 1124AD in the siege of Tyre (in present-day Lebanon) during the Crusades. As it was much more powerful than a catapult, a trebuchet became the siege weapon of choice.

Explanation:

Most people think that the trebuchet is a medieval weapon, but actually, it has its origins in ancient China. The ancient Chinese trebuchet could throw huge boulders up to 250 pounds, a distance of 200 feet. In addition, it was lightweight and mobile which was crucial as it could be moved around the battlefield with ease. The thing I love about the chinese trebuchet is that it pushes the boundaries of ancient engineering to create a ballistics revolution - a flexible, portable, killing machine.

The total voltage of the circuit is the product of the total current and resistance through the circuit. Hence, the voltage here is 2.17 ×10⁻⁹ V.

Ohm's law states that, the total voltage across a circuit is the product of the current over the circuit and the overall resistance. Let V be the voltage , I and R be the current and resistance respectively.

then,

V = IR.

Given,

R = 3.4 milli Ω = 3.4 × 10⁻³ Ω

I = 0.16 micro amperes = 0.16 × 10⁻⁶ A.

Then voltage V = IR =  0.16 × 10⁻⁶ A ×  3.4 × 10⁻³ Ω

V = 5.4 ×10⁻¹⁰ V.

This is the voltage required for one system. For all the 4 systems,

v  = 4 × 5.4 ×10⁻¹⁰ V = 2.17 ×10⁻⁹ V.

Therefore, the total voltage needed to supply the systems is 2.17 ×10⁻⁹ V.

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Answer:

Longitudinal waves

Explanation:

With sound waves, the energy travels along in the same direction as the particles vibrate. This type of wave is known as a longitudinal wave , so named because the energy travels along the direction of vibration of the particles.

Answer:

895522 times faster.

Explanation:

From the question given above, the following data were obtained:

Speed of sound in air (v) = 335 m/s

Speed of light in air (c) = 3×10⁸ m/s

How many times faster =.?

To obtain how many times faster light travels in air than sound, do the following

c : v => 3×10⁸ : 335

c/v = 3×10⁸ / 335

c/v = 895522

Cross multiply

c = 895522 × v

From the illustrations made above, we can see that the speed of the light in air (c) is 895522 times the speed of sound in air.

Thus, light travels 895522 times faster than sound in air.

Answer: 1.232

Explanation:I remember having this as a hw and this was the answer correct me if I was wrong

From Hooke's law, the length of the spring after extension and after the mass is attached is 1.788 meters. Option B is the answer.

Hooke's law state that in an elastic material, the force applied is directly proportional to the extension provided that the elastic limit is not exceeded.

Given that a 12.0 kg mass is attached to 1.2 m long spring with a spring constant of 200.0 N/m.

The given parameters are:

According to Hooke's law

F = Ke

But F = mg

mg = Ke

Substitute all the parameters into the formula to get extension e

12 x 9.8 = 200e

e = 117.6 / 200

e = 0.588 m

The length of the spring after extension and after the mass is attached will be

\(L_{2}\) = \(L_{1}\) + e

\(L_{2}\) = 1.2 + 0.588

\(L_{2}\) = 1.788 m

Therefore, the correct answer is option B because the length of the spring after extension and after the mass is attached is 1.788 meters.

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In the limit of large w (w ≥ 1/RC, 1/√(LC), R/L), the statement "Vout is approximately equal to Vin" is true.

When the frequency w is large, the reactance of the capacitor (1/wC) and the inductor (wL) become significant. In this limit, we can analyze the circuit using impedance concepts.

The impedance of the series LRC circuit is given by Z = R + j(wL - 1/wC), where j is the imaginary unit. The magnitude of the impedance is |Z| = sqrt(R^2 + (wL - 1/wC)^2).

In the limit of large w, the term 1/wC dominates the impedance, making the magnitude of Z approximately equal to R. Therefore, the voltage drop across the resistor dominates, and Vout becomes approximately equal to Vin.

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The effective value of g, the acceleration of gravity, at 6700m , above the Earth's surface is 9.79 m/s².

Given:

G=gravitational constant = 6.67x10⁻¹¹ m³/(kg.s²)

M=the Earth's mass = 5.97x10²⁴ kg

r=  Earth's radius = 6371 km  +6700m

r=6371000m+6700m

r=6377700m

For gravitation we use following:

F=GmM/r^2

which on further derivation gives

ma=GmM/r^2 , which results in

a=GM/r^2

Now putting the values in a=GM/r^2

a=6.67x10⁻¹¹ m³/(kg.s²)*5.97x10²⁴ kg/6377700^2

a=9.79 m/s²

Hence, The effective value of g, the acceleration of gravity, at 6700m , above the Earth's surface is 9.79 m/s².

then at 7000km i.e. 7000000m

All will be same except r= 6371000m+7000000m

r=13371000m and

r^2 =1.78*10^14.

putting values

a=6.67x10⁻¹¹ m³/(kg.s²)*5.97x10²⁴ kg/1.78*10^14.

a=2.23*10^28 m/s.

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Answer:

true

Explanation:

basic sports knowlegde

Answer:

4.2 N. S

Explanation:

Impulse =Force X Time

                35 X 0.12

                 4.2 N

Answer:

0.39 J/g°c

Explanation:

= heat / unit of mass × unit of temperature

986.75J/16.75g

= 58.9 J/g

∆T=175°c - 25°c = 150°c

986.75 / 150°c = 6.578

986.75 / 16.75g.150°c = 0.30 j/g°c

Answer:

Water gains energy during evaporation and releases it during condensation in the atmosphere

Explanation:

In the water cycle, heat energy is gained or lost by water as it undergoes various processes in the cycle.

In evaporation, water molecules gains energy because the molecules of water vibrate faster and become more energetic. Hence they are able to escape into the atmosphere from the surface of the liquid.

In condensation, the molecules of gaseous water looses energy and becomes liquid.

Hence, water gains energy during evaporation and releases it during condensation in the atmosphere.

Answer:

K12 HE HE

Explanation:

The ball stays in the air for a total of 1.5 + 0 = 1.5 seconds. The total time the ball stays in the air from the time it was thrown until it returns to the point of release can be calculated by considering the time it takes to reach the highest point and the time it takes to fall back down.

1. The total time the ball stays in the air from the time it was thrown until it returns to the point of release can be calculated by considering the time it takes to reach the highest point and the time it takes to fall back down.
Given that the ball reached the highest point after 3 seconds, we can assume that it took 1.5 seconds to reach the highest point. This is because the time taken to reach the highest point is half of the total time in the air.
To calculate the time it takes for the ball to fall back down, we can use the equation:
t = sqrt((2h) / g)
Where t is the time, h is the height, and g is the acceleration due to gravity (approximately 9.8 m/s^2). Since the ball has returned to the point of release, the height is zero.
Plugging in the values, we have:
t = sqrt((2 * 0) / 9.8) = 0 seconds
Therefore, the ball stays in the air for a total of 1.5 + 0 = 1.5 seconds.
2. The final velocity of the ball when it returns to the point of release can be determined by considering the initial velocity and the acceleration due to gravity.
When the ball is thrown upward, the initial velocity is 29.4 m/s. As the ball reaches the highest point, its velocity becomes zero. When the ball falls back down, it accelerates due to gravity and gains velocity.
The final velocity can be calculated using the equation:
v = u + gt
Where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken to reach the highest point (1.5 seconds).
Plugging in the values, we have:
v = 29.4 + (9.8 * 1.5) = 29.4 + 14.7 = 44.1 m/s
Therefore, the final velocity of the ball when it returns to the point of release is 44.1 m/s.
To summarize, the ball stays in the air for 1.5 seconds from the time it was thrown until it returns to the point of release. The final velocity of the ball when it returns to the point of release is 44.1 m/s.

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In a fire piston, the temperature of the gas inside the piston increases due to adiabatic compression.

In physics, pressure is a measure of the force per unit area that a fluid (liquid or gas) exerts on its surroundings. It is defined as the force F acting perpendicular to a surface divided by the area A over which the force is distributed: pressure = F/A. The SI unit of pressure is the Pascal (Pa), which is defined as one Newton per square meter (N/m²). Other common units of pressure include pounds per square inch (psi), atmospheres (atm), and bar (bar). Pressure is an important concept in many areas of physics, including fluid mechanics, thermodynamics, and atmospheric science. It describes the behavior of fluids in pipes and channels, the operation of hydraulic systems, and the behavior of gases in sealed containers. The pressure of a fluid is related to its density and temperature, and can be influenced by factors such as gravity, elevation, and the motion of the fluid itself.

Here,

When the piston is rapidly pushed down into the cylinder, it compresses the gas inside the cylinder. This rapid compression causes the gas molecules to collide with each other more frequently and with greater force, which in turn increases the temperature of the gas.

Because the compression is adiabatic (i.e., no heat is allowed to enter or leave the system), the increase in temperature is due solely to the work done on the gas by the piston. This increase in temperature is known as adiabatic heating.

The fire piston is a simple but effective tool for starting a fire in the wilderness, and it demonstrates the principles of adiabatic compression and the relationship between pressure, volume, and temperature in a gas.

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Answer:

4 days

either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2

n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

The half life of the sample will be "4 days".

According to the question,

Let,

→  \(\frac{N}{N_o} = (\frac{1}{2} )^n\)

By substituting the values, we get

→  \(\frac{2}{128} = (\frac{1}{2} )^n\)

→    \(\frac{1}{64} = (\frac{1}{2} )^n\)

→ \((\frac{1}{2} )^6= (\frac{1}{2} )^n\)

→     \(n=6\)

So,

It takes 6 half life just to reduce from 128 mg - 2mg which is also "24 days".

→ \(6 \ half \ life = 24 \ days\)

Hence,

→ \(Half \ life = \frac{24}{6}\)

                   \(= 4 \ days\)

Thus the answer above is right.

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The power required to do all the work above is 2756.25 J / 20.5 s = 134.3 W.

To lift the 22.5 kg bag of trash 2.5 m off the ground, you did work against gravity. The work done can be calculated using the formula W = mgh, where m is the mass of the object, g is the acceleration due to gravity (\(9.8 m/s^2\)), and h is the height lifted. Therefore, W = \((22.5 kg)(9.8 m/s^2)(2.5 m)\) = 551.25 J.
To carry the bag of trash up a flight of stairs that are 10 m tall, you did work against gravity as well. The work done can be calculated using the same formula. Therefore, W = \((22.5 kg)(9.8 m/s^2)(10 m)\) = 2205 J.
The total work done on the bag of trash is the sum of the work done to lift it off the ground and the work done to carry it up the stairs. Therefore, the total work done is 2756.25 J.
To calculate the power required to do all the work above, we can use the formula P = W/t, where P is power, W is work done, and t is time.

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Answer:

product

Explanation: it's the product of chlorine and sodium

Answer:

y

Explanation:

A regular polygon's interior angles add up to 3060° of a 19 sided polygons.

Polygons are defined as a flat shape surrounded by sides, or line segments.

It can be defined as a plane figure whose description is given by a closed polygonal chain made up of a finite number of straight line segments.

It can also be defined as any closed curve made up of a number of connected line segments that do not cross one another.

Sum of interior angle = (p - 2) x 180°

Where p = 19

Sum of interior angle = (19 - 2) x 180°

                                   = 17 x 180°

                                   = 3060°

Thus, a regular polygon's interior angles add up to 3060° of a 19 sided polygons.  

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Answer:

B

Explanation:

Answer:

B

Explanation:

I had a feeling

The critical angle is the angle of incidence that results in the refracted ray traveling along the interface between two different materials, in this case water and fused quartz.

To find the critical angle, we can use Snell's law, which relates the angle of incidence to the refractive indices of the materials involved.
Snell's law states: \(n1 * sin(theta1) = n2 * sin(theta2)\)

Here, n1 is the refractive index of the first medium (water), theta1 is the angle of incidence, n2 is the refractive index of the second medium (fused quartz), and theta2 is the angle of refraction.

In this case, we have n1 = 1.33 (refractive index of water) and n2 = 1.46 (refractive index of fused quartz). To find the critical angle, we need to determine the angle of incidence (theta1) that will result in the refracted ray traveling along the interface.

Using Snell's law, we can rewrite it as: \(sin(theta1) = (n2/n1) * sin(theta2)\)

For the critical angle, sin(theta2) will be equal to 1 (as the refracted ray will be at a 90-degree angle with the normal). Therefore, we have: \(sin(theta1) = (n2/n1)\)

Plugging in the values, we get: \(sin(theta1) = (1.46/1.33)\)

To find theta1, we can take the inverse sine of both sides:\(theta1 = sin^(-1)((1.46/1.33))\)

Using a calculator, we find theta1 ≈ 43.76 degrees.

Therefore, the critical angle for the interface between water and fused quartz is approximately 43.76 degrees.

The critical angle for the interface between water and fused quartz is approximately 43.76 degrees.

The critical angle is the angle of incidence at which the refracted ray no longer passes into the second medium, but rather travels along the interface between the two materials.

To find the critical angle, we can use Snell's law, which relates the angle of incidence to the refractive indices of the materials involved.

By plugging in the given values, n1 = 1.33 (refractive index of water)

and n2 = 1.46 (refractive index of fused quartz), we can use the equation \(sin(theta1) = (n2/n1)\) to find the angle of incidence.

Taking the inverse sine of both sides, we find that the critical angle is approximately 43.76 degrees.

The critical angle for the interface between water and fused quartz is approximately 43.76 degrees.

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Answer:

facts provided or learned about something or someone.

Answer:

100 N

Explanation:

For ONE bullet ( .025 kg)

 fired to 1000 m/s   in  .25 s   ( four per second)

     acceleration = 1000 m/s   /  .25 s = 4000 m/s^2

then F = ma

           = .025 kg * 4000 m/s^2 = 100 N

Answer:

False

Explanation:

.......................

The image given, is explained in terms of Newton's 1st law of motion.

Based on Newton's First Law of Motion (Inertia).  An object at rest remains at rest, and an object in motion continues in motion at a steady speed and in a straight path until operated on by an unbalanced force.

Newton's first rule of motion states that if a pan of water were being moved along a track, the water would tend to keep moving ahead. However, the water will appear to splash to the right while it goes to the left. Space is nearly a perfect vacuum, devoid of both gravity and matter. Think about a satellite travelling 17,500 mph around the Earth. If a rock were to be launched from the satellite, it would orbit the earth adjacent to the satellite at a speed of 17,500 mph.

Here, in the given image, it is shown that the sofa is remain in the place unless a person forcefully move it from one place to another.

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An example of no work after the effort is work done by the collie to travel a distance in a straight line by lifting a mass.

In physics, the word "work" involves the measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied within the direction of the displacement.

The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.

The amount of work done by the coolie to travel a certain distance in a straight line by lifting a certain amount of mass. Here the work done is zero because when the coolie lifted the mass the force exerted in the upward direction makes an angle of 90° and as per the formula of work done it will be equal to zero.

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Properly bonding all metal parts within an electrical system helps ensure a low-impedance fault current path. Fault current is the current that flows when an electrical system experiences a fault or short circuit.

This current can cause damage to equipment and pose a risk of electric shock to personnel if not properly controlled. A low-impedance fault current path is necessary to ensure that fault current is safely and quickly conducted away from the point of the fault.To achieve a low-impedance fault current path, all metal parts within an electrical system must be bonded together using suitable conductors or connectors. This includes the equipment grounding conductor (EGC), which is a dedicated conductor that connects all non-current-carrying metal parts of electrical equipment together and to the system grounding conductor (SGC).

The SGC provides a connection to earth and ensures that any fault current is safely directed away from the system to ground. By bonding all metal parts together, electrical continuity is established throughout the system, reducing the risk of high-impedance or intermittent connections that can lead to arcing, overheating, and other hazards. Proper bonding also reduces electromagnetic interference (EMI) and ensures the effectiveness of overcurrent protection devices, such as fuses and circuit breakers.

Overall, proper bonding is an essential element of electrical safety, ensuring that fault current is quickly and safely conducted away from the point of the fault, reducing the risk of electrical hazards and equipment damage.

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