Design a circuit which will correct a load of 180kW at 0.85 lagging power factor to 0.97 lagging power factor. Assume that the load is supplied by a 220V (rms), 50 Hz line. 4.364mF 14.671mF 8.728mF 7.335mF 1000

6.00 g bullet, traveling horizontally at a speed of 400 m/s  is initially at rest on a horizontal surface then moment conservation V=0.4 m/s

Mass bullet = 6*10⁻³

Mass of block=2kg

Pi=Pf

(5*10⁻³)400+0=120*(6*10⁻³)+2v

v=0.7 m/s

Ki=1/2mv²=1/2ˣ6*10⁻³ˣ400²

Ki=1/2ˣ6ˣ10⁻³ˣ16ˣ10⁴

Ki=400j

Kf=1/2mv²=1/2ˣ6*10⁻³ˣ120²

Kf=36J

moment conservation

V=0.4 m/s.

The principle of momentum conservation states that momentum is neither created nor destroyed but only changed by the action of forces as they are described by Newton's laws of motion. This applies to a particular problem domain. The only thing that happens when two bodies collide is that their momentum changes. Thus, for bodies colliding, changes in momentum are always equal and opposite. The momentum of the other body must decrease by the same amount if the momentum of the first body increases. As a result, momentum always remains constant.

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Answer:

C) 10000 J

Explanation:

∆p.e =mg∆h

= 2000 × 5

= 10000J

Answer:

The Pauling-Branson Award recognizes the contributions of Herman Russell Branson, (1914-1995), one of the first African American physicists to make crystallography the focus of his research.

Answer:

Increase height

Increase mass

The properties that can be observed from the use of telescope and other instruments include colour, parallax angle, spectral type, apparent brightness. The things that are inferred indirectly are luminosity, surface temperature, mass and radius.

A binary star is a system of two stars that are gravitationally bound to orbit around each other. Binary system in the night sky that are seen as a single object using the eye are often resolved using a telescope as separate stars known as visual binaries.

However, we can measure a star's mass directly if it is a member of an eclipsing binary system.  

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On which planet would a 15.0 kg object weigh the most?The weight of an object is affected by its mass and the strength of gravity. On planets with a greater mass, the force of gravity is stronger than on those with a lower mass. Hence, the weight of an object is greater on planets with more mass, according to Newton's second law of motion.

The weight of a 15 kg object is the largest on the planet Jupiter because Jupiter has the highest mass among all the planets of our solar system. The answer is written below:Jupiter is the planet where a 15.0 kg object would weigh the most.Round the answer to the nearest whole numberOn Jupiter, the acceleration due to gravity is 24.79 m/s². The weight of an object on Jupiter is calculated using the formula:

F = m × gwhereF is the force of gravity (in newtons),m is the mass of the object (in kilograms), andg is the acceleration due to gravity (in meters per second squared).On Jupiter, the weight of a 15.0 kg object is:F = 15.0 kg × 24.79 m/s²= 371.85 N = 372 N (to the nearest whole number)Therefore, on Jupiter, a 15.0 kg object would weigh about 372 N.

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Answer:

2500 Newtons

Explanation:

If force equals the mass × acceleration then all should be quite simple

25kg is the mass of your car

100 m/s/s is the acceleration of you car

25kg • 100m/s/s = 2500 Newtons

Answer:

\(\boxed {\boxed {\sf 2500 \ Newtons }}\)

Explanation:

The formula for force is the product of mass and acceleration.

\(F=m*a\)

The mass of the car is 25 kilograms and the acceleration is 100 meters per square second. So,

\(m= 25 \ kg \\a= 100 \ m/s^2\)

Substitute the values into the formula.

\(F= 25 \ kg * 100 \ m/s^2\)

Multiply.

\(F= 2500 \ kg*m/s^2\)

\(F= 2500 \ N\)

The force of the car is 2500 Newtons.

Answer:

Science can not solve all of our problems. While scientific understanding can help battle things like disease, hunger, and poverty when applied properly, it does not do so completely and automatically. Furthermore, there are many areas of life where science can have little impact. Let us look at some of the reasons why this is so.  First of all, there is a huge difference between knowing something and acting on it. Science is concerned with accumulating and understanding observations of the physical world. That understanding alone solves no problems. Individual people have to act on that understanding for it to help solve problems. For instance, science has found that regular exercise can lower your risk of heart disease. Knowing this fact is interesting, but it will do nothing for your personal heath unless you act on it and actually exercise. And that's the hard part. Reading an article about exercise is easy. Getting into an actual routine of regular exercise is harder. In this sense, science really solves no problems at all. Problems are only solved when people take the knowledge (or tool, or pill, or whatever) provided by science and use it. In fact, many of humanity's biggest problems are caused by lack of action, and not lack of knowledge.

Take world hunger, for example. There is currently enough food produced on the earth every year to comfortably feed every single person. The world produces about 700 trillion grams of rice each year. With seven billion people on the planet, 365 days in the year, and about 40 grams per typical serving of rice, there is enough rice on the planet to feed every single last person seven servings of rice every day. And this is just rice. Similar numbers hold up for wheat, corn, meat, etc. Science has done an amazing job in the last 50 years of making farms productive. And yet, millions of people in the world still suffer starvation. Why? Because of actions. If all it took was science to solve problems, no one would go hungry anymore because there is enough food. We could fill books with the analysis of human actions that cause world hunger if we wanted to.

Explanation:

Only queries that can be critiqued through observation and experiment are investigated using the scientific method. As a result, science cannot answer value-laden questions like the meaning of life.

Science is the study of the natural world. This includes the physical universe's components, such as atoms, plants, ecosystems, people, societies, and galaxies, in addition to the innate forces acting on those things.

Science, on the other hand, cannot investigate supernatural forces and explanations.

Environmental science aids in our understanding of the natural world because it encompasses all aspects of nature.

The natural world is comprised of numerous components, including the Earth itself, our atmosphere, bodies of water, and all living things that inhabit our planet.

The scientific method is used to investigate only questions that can be challenged through observation and experimentation.

Thus, scientist study the natural world to find solutions to human problems.

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A  minimum coating thickness of 10.35 cm would be required to render an aircraft invisible to a radar operating at 725 MHz.

In order to determine the minimum thickness of coating required to render an aircraft invisible to a radar operating at 725 MHz, we first need to know the wavelength of the radar signal. The wavelength can be calculated using the formula:

wavelength = speed of light / frequency

The speed of light is approximately 3 x 10^8 meters per second. Converting the frequency of 725 MHz to meters, we get:

wavelength = 3 x 10^8 / (725 x 10^6) = 0.4138 meters

Now, in order to render the aircraft invisible to this radar band, we need the coating thickness to be at least one-quarter of the wavelength. Therefore, the minimum thickness of the coating required would be:

minimum coating thickness = 0.4138 meters / 4 = 0.1035 meters

Converting the thickness to centimeters:

minimum coating thickness = 10.35 cm

Therefore, a minimum coating thickness of 10.35 cm would be required to render an aircraft invisible to a radar operating at 725 MHz.

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Answer:

Explanation:

Bigger pushes and pulls cause bigger changes in an object's motion or shape. When objects collide, contact forces transfer energy so as to change the objects' motions. When two objects interact, each one exerts a force on the other, and these forces can transfer energy between them.

To answer your question, let's break it down into steps.

1. Calculate the initial velocity:

2. Calculate the final velocity:

3. Calculate the change in velocity:

4. Calculate the time it takes to slow down:

5. Use the equation of motion to find the time:

6. Calculate the acceleration:

7. Substitute Δt into the equation from step 5:

8. Solve for a:

9. Calculate the acceleration:

10. Calculate the time to slow down:

Speed ​​skater  is a form of competitive ice skating in which competitors race each other over a certain distance on roller skates. Types of speed skating are long track speed skating, short track speed skating, and marathon speed skating. In the Olympics, long track speed skating is usually referred to simply as "sprinting", while short track speed skating is known as "short track". The International Skate Union (ISU), the governing body for both esports, refers to the long track as "speed skating" and the short course as "short course skating".

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Answer:

\(a_1 = 1.446m/s^2\)

Explanation:

Given

\(m_1 = 70.0kg\) -- Mass of the boy

\(m_2 = 45.0kg\) -- Mass of the girl

\(a_2 = 2.25m/s^2\) -- Acceleration of the girl towards the boy

Required

Determine the acceleration of the boy towards the girl (\(a_1\))

From the question, we understand that the surface is frictionless. This implies that the system is internal and the relationship between the given and required parameters is:

\(m_1 * a_1 = m_2 * a_2\)

Substitute values for \(m_1, m_2\) and \(a_2\)

\(70.0 * a_1 = 45.0 * 2.25\)

Make \(a_1\) the subject

\(\frac{70.0 * a_1}{70.0} = \frac{45.0 * 2.25}{70.0}\)

\(a_1 = \frac{45.0 * 2.25}{70.0}\)

\(a_1 = \frac{101.25}{70.0}\)

\(a_1 = 1.446m/s^2\)

Hence, the acceleration of the boy towards the girl is 1.446m/s^2

The term you are looking for is "chemical battery". Chemical batteries work by converting chemical energy into electrical energy through a series of chemical reactions. These reactions take place within the battery's cells, which are composed of two electrodes and an electrolyte.

When the battery is connected to a circuit, the chemical reactions produce an electrical current that can be used to power devices. Chemical batteries are widely used in many applications, including consumer electronics, electric vehicles, and renewable energy systems. They are a crucial component of our modern technological society, and ongoing research is focused on developing more efficient and sustainable battery technologies to meet growing energy demands.

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Answer:a licensed physician

Explanation: taking the osha test now.

A licensed physician is the one to carry out a post-exposure medical evaluation. (Option A)

Post-exposure Medical Evaluation is any medical evaluation started after exposure to a pathogen in order to prevent the infection from occurring.

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The trip takes a total of 1.5 hours and has an average speed of 40 mph.

To calculate the time taken for each leg of the trip, we can use the formula time = distance/speed.

For the first leg of the trip, the car travels 15 miles at a speed of 45 mph. Using the formula, we find that the time taken for this leg is 15/45 = 0.33 hours.

For the second leg of the trip, the car travels another 15 miles but at a speed of 30 mph. Using the formula, we find that the time taken for this leg is 15/30 = 0.5 hours.

To find the total time for the trip, we add the times for each leg: 0.33 hours + 0.5 hours = 0.83 hours.

To calculate the average speed for the entire trip, we use the formula average speed = total distance/total time. The total distance traveled is 15 miles + 15 miles = 30 miles. The total time taken is 0.83 hours. Plugging these values into the formula, we find that the average speed for the trip is 30/0.83 = 36.14 mph.

Therefore, the trip takes a total of 1.5 hours and has an average speed of 40 mph.

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This is due to the fact that occluded fronts combine two cold air masses, which causes one of the cold air masses to go downward.

When a warm air mass is sandwiched between two cold air masses, an occluded front occurs. In an occlusion, the warm front passes over the cold front, which dives beneath it.

In a front is obscured, the warm front is fully supplanted by the cold front, in which the warm air masses have completely disappeared. Furthermore, there are frequent shifts in the various weather producing circumstances because of the cold front's relatively low temperature.

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A 0 20-kg hockey puck initially traveling 5.0 m/s is pushed on ice by a hockey stick with a constant force of 7.0 N over a distance of 8.0 m. Friction is negligible. The final velocity of the hockey puck can be determined by using the equation, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the distance.Vf² = Vi² + 2ad

The force exerted on the hockey puck is given as,F = 7.0 NThe distance travelled by the hockey puck is given as,d = 8.0 mThe initial velocity of the hockey puck is given as,Vi = 5.0 m/sThe mass of the hockey puck is given as,m = 0.20 kgWe can determine the acceleration experienced by the hockey puck by using the following formula,

F = maSolving for a, we geta = F / ma = F / ma = (7.0 N) / (0.20 kg)a = 35.0 m/s²Substituting the values in the formula for final velocity,Vf² = Vi² + 2adVf² = (5.0 m/s)² + 2 × (35.0 m/s²) × (8.0 m)Vf² = 25.0 m²/s² + 560.0 m²/s²Vf² = 585.0 m²/s²Taking the square root of both sides, we get,Vf = √(585.0 m²/s²)Vf = 24.2 m/sTherefore, the final velocity of the hockey puck is 24.2 m/s.

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A real root of the equation f(x) = x^2 - 2x - 5 = 0, found using the bisection method in five stages, is approximately x = -0.046875. To find a real root of the equation f(x) = x^2 - 2x - 5 = 0 using the bisection method, we can start by identifying an interval [a, b] that contains the root. Let's choose the interval [-3, 0], where f(a) = f(-3) = 4 and f(b) = f(0) = -5. Since f(a) and f(b) have opposite signs, there must be a root within this interval.

Stage 1:

- Start with interval [a, b] = [-3, 0]

- Calculate the midpoint c = (a + b) / 2 = (-3 + 0) / 2 = -1.5

- Evaluate f(c) = (-1.5)^2 - 2(-1.5) - 5 = -0.25

- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-1.5, 0]

Stage 2:

- Calculate the new midpoint c = (a + b) / 2 = (-1.5 + 0) / 2 = -0.75

- Evaluate f(c) = (-0.75)^2 - 2(-0.75) - 5 = -4.4375

- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-0.75, 0]

Stage 3:

- Calculate the new midpoint c = (a + b) / 2 = (-0.75 + 0) / 2 = -0.375

- Evaluate f(c) = (-0.375)^2 - 2(-0.375) - 5 = -2.7461

- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-0.375, 0]

Stage 4:

- Calculate the new midpoint c = (a + b) / 2 = (-0.375 + 0) / 2 = -0.1875

- Evaluate f(c) = (-0.1875)^2 - 2(-0.1875) - 5 = -1.2217

- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-0.1875, 0]

Stage 5:

- Calculate the new midpoint c = (a + b) / 2 = (-0.1875 + 0) / 2 = -0.09375

- Evaluate f(c) = (-0.09375)^2 - 2(-0.09375) - 5 = -0.6104

- Since f(c) has the same sign as f(a), replace a with c: [a, b] = [-0.09375, 0]

After five stages, the interval [a, b] has become [-0.09375, 0]. Since the interval is small, we can approximate the root as the midpoint of this interval:

Root ≈ (a + b) / 2 = (-0.09375 + 0) / 2 = -0.046875

Therefore, a real root of the equation f(x) = x^2 - 2x - 5 = 0, found using the bisection method in five stages, is approximately x = -0.046875.

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As you walk away from the charge, the potential becomes less negative and actually grows, coming closer and closer to zero.

If you are infinitely far away from the charge, the potential for both positive and negative charges is zero. The electric potential (also known as the electric field potential, potential drop, or electrostatic potential) is defined as the amount of work energy required to transport a unit of electric charge in an electric field from a reference point to a specified place. The Volt difference dropped across a fixed resistance of one ohm with a current of one ampere running through it is the unit of potential difference formed between two locations.

dV

​V=5x \s2 \s −10x−9

dx dV \s​ \s

=10x−10

so, E=−(10x−10)

The value of E=(1010)

=0V/m for

x=1m

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Answer:

the balloon is filled with the 8.5 litre of helium. temperature is 294 kelvin.

party starts at 4 p.m.

the temperature rises 305 kelvin.

the new volume = 4 litres

Answer: 5

Explanation:

Given information: The average method reports the following unit data for the forming department. Units completed in the forming department are transferred to the painting department. Production cost information for the forming department follows.

Direct materials:
Units completed during the period = 45,000 units
Ending work in process inventory = 5,000 units
Direct materials cost = $202,500

Conversion costs:
Units completed during the period = 45,000 units
Ending work in process inventory = 5,000 units
Conversion cost = $189,000

a. Calculation of equivalent units of production for both direct materials and conversion for the forming department:
Equivalent units of production = Units completed during the period + (Ending work in process inventory * Degree of completion)
Direct materials:
Equivalent units of production = 45,000 + (5,000 * 50%) = 47,500 units

Conversion costs:
Equivalent units of production = 45,000 + (5,000 * 60%) = 48,000 units

b. Calculation of the cost per equivalent unit of production for both direct materials and conversion for the forming department:
Cost per equivalent unit of production = Total cost for the period / Equivalent units of production

Direct materials:
Cost per equivalent unit of production = $202,500 / 47,500 units = $4.26 per unit

Conversion costs:
Cost per equivalent unit of production = $189,000 / 48,000 units = $3.94 per unit

c. Calculation of the cost assigned to the forming department's output using the weighted average method:
Total cost = Cost of units transferred out + Cost of ending work in process inventory
Cost of units transferred out = Number of units transferred out * Cost per equivalent unit of production
Cost of ending work in process inventory = Number of units in ending work in process inventory * Cost per equivalent unit of production

Direct materials:
Cost of units transferred out = 40,000 * $4.26 per unit = $170,400
Cost of ending work in process inventory = 5,000 * $4.26 per unit = $21,300
Total cost = $170,400 + $21,300 = $191,700

Conversion costs:
Cost of units transferred out = 40,000 * $3.94 per unit = $157,600
Cost of ending work in process inventory = 5,000 * $3.94 per unit = $19,700
Total cost = $157,600 + $19,700 = $177,300

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The Minimum Angular Separation Between Two Stars is 4 × 10−6 Rad, If Telescope is Used to Observe Them with an Objective of Aperture 16 Cm.

The angular resolution measure is the angular size of the smallest features that the telescope can see. θ ≈ a / d . A complete circle is 2π radians = 360 degrees. Angular sizes or separations may also be measured in arcminutes (60 arcmin = 1 degree) or arcseconds (60 arcsec = 1 arcmin).

To find Angular Separation

It represents cosine angle between two vectors. ...

Formula. ...

Exercise: ...

From vector algebra we remember that cosine angle between two vectors can be represented as dot product divided by length of the two vectors.

The length of a vector (sometimes called modulus) is the root of square of its coordinate.

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The distance between point O and point A is √19. The distance between point O and point S is √61. The distance between the ant and the spider is √34.

1. To find the distance between point O and point A, we can use the distance formula in three-dimensional space. The distance formula is:

d = √((x2 - x1)² + (y2 - y1)² + (z2 - z1)² )

Substituting the coordinates of point O (1, 0, 2) and point A (2, 3, 5) into the formula, we have:

d = √((2 - 1)²  + (3 - 0)²  + (5 - 2)² )
 = √(1²  + 3²  + 3² )
 = √(1 + 9 + 9)
 = √19

Therefore, the distance between point O and point A is √19.

To find the distance between point O and point S, we can follow the same steps. Substituting the coordinates of point O (1, 0, 2) and point S (6, 0, 8) into the distance formula, we have:

d = √((6 - 1)²  + (0 - 0)²  + (8 - 2)² )
 = √(5²  + 0 + 6² )
 = √(25 + 0 + 36)
 = √61

Therefore, the distance between point O and point S is √61.

2. To find the distance between the ant and the spider, we can use the distance formula once again. Substituting the coordinates of point A (2, 3, 5) and point S (6, 0, 8) into the formula, we have:

d = √((6 - 2)²  + (0 - 3)²  + (8 - 5)² )
 = √(4²  + (-3)²  + 3² )
 = √(16 + 9 + 9)
 = √34

Therefore, the distance between the ant and the spider is √34.

In conclusion,
1. The distance between point O and point A is √19.
2. The distance between point O and point S is √61.
3. The distance between the ant and the spider is √34.

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To determine the electric field outside a charged sphere, we can use Gauss's Law, which states that the flux of the electric field through a closed surface is proportional to the charge enclosed by the surface.

If the charge is distributed uniformly throughout the spherical volume of radius R, then the charge enclosed by a Gaussian surface outside the sphere is simply the total charge of the sphere.

The electric field outside the sphere is radial, and by symmetry, it must have the same magnitude at any point on a sphere with radius r greater than R.

Therefore, we can choose a spherical Gaussian surface with radius r > R and calculate the flux of the electric field through that surface.

By Gauss's Law, the flux is proportional to the charge enclosed, which is the total charge Q of the sphere.

The electric field magnitude E is related to the flux Φ and the surface area A of the Gaussian surface by:

Φ = E * A

where Φ = Q / ε_0 is the total electric flux through the surface, and ε_0 is the electric constant.

Since the Gaussian surface is a sphere, its surface area is 4πr^2. Therefore, we have:

E * 4πr^2 = Q / ε_0

Solving for E, we get:

E = Q / (4πε_0r^2)

This expression tells us that the electric field outside the sphere decreases with the square of the distance from the center of the sphere.

At large distances (compared to the radius of the sphere), the electric field can be approximated as that of a point charge, with the same total charge as the sphere, located at the center of the sphere.

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The melting ice at the same temperature will have greater internal energy as compared to the ice at 0° C.

In thermodynamics, the internal energy is a quantity or state function that characterizes a substance's energy when capillary effects and other external fields, such as electric and magnetic forces, are absent. The value of the energy depends on the substance's state, not the nature of the processes that brought it to that condition, just like any other state function.

The melting ice at the same temperature will have more internal energy as compared to the ice at 0° C because during the melting of ice the intermolecular particles will absorb energy while breaking their bond and because of this, the melting ice has greater kinetic energy as well.

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The Rankine-Hugoniot jump conditions, we get: \(γ * ρ2 * 420 m/s\)

To solve this problem, we can use the Rankine-Hugoniot jump conditions, which relate the properties of a fluid across a shock wave:

\(ρ1 * v1 = ρ2 * v2p1 + ρ1 * v1^2 = p2 + ρ2 * v2^2\)

where ρ is the density, v is the velocity, and p is the pressure.

Let's assume that the shock waves are traveling in the x-direction. We can define a coordinate system where the shock waves intersect at x=0, and the first shock wave is moving to the right, while the second shock wave is 10 m behind and also moving to the right.

Before the shocks intersect, the state of the gas is uniform and can be described by the properties of air at standard atmospheric conditions. We can assume that the gas is at rest, so v1 = v2 = 0.

Let's first find the properties of the gas behind the second shock wave. We can use the Rankine-Hugoniot jump conditions to relate the state behind the second shock wave (denoted by subscript "2") to the standard atmospheric state (denoted by subscript "0"):

\(ρ2 * v2 = ρ0 * v0p2 + ρ2 * v2^2 = p0 + ρ0 * v0^2\)

Since the second shock wave is traveling at 840 m/s, the velocity behind it is also 840 m/s:

v2 = 840 m/s

We can assume that the density behind the second shock wave is higher than the standard atmospheric density because the shock wave compresses the gas. Let's define the density behind the second shock wave as ρ2 = α * ρ0, where α is the density ratio. Similarly, let's define the pressure behind the second shock wave as p2 = β * p0, where β is the pressure ratio.

Substituting these expressions into the Rankine-Hugoniot jump conditions, we get:

\(α * ρ0 * 840 m/s = ρ0 * 0 m/sβ * p0 + α * ρ0 * (840 m/s)^2 = p0 + ρ0 * 0^2\)

Solving for α and β, we get:

α = 0

β = 2.25

This means that the density behind the second shock wave is zero, and the pressure is 2.25 times the standard atmospheric pressure.

Now let's find the properties of the gas behind the first shock wave. We can use the Rankine-Hugoniot jump conditions to relate the state behind the first shock wave (denoted by subscript "1") to the state behind the second shock wave (denoted by subscript "2"):

\(ρ1 * v1 = ρ2 * v2p1 + ρ1 * v1^2 = p2 + ρ2 * v2^2\)

Since the first shock wave is traveling at 420 m/s, the velocity behind it is also 420 m/s:

v1 = 420 m/s

We can assume that the density behind the first shock wave is higher than the density behind the second shock wave because the first shock wave compresses the gas further. Let's define the density behind the first shock wave as\(ρ1 = γ * ρ2,\) where γ is the density ratio. Similarly, let's define the pressure behind the first shock wave as p1 = δ * p2, where δ is the pressure ratio.

Substituting these expressions into the Rankine-Hugoniot jump conditions, we get:

γ * ρ2 * 420 m/s

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The radial force on the gear is approximately 5041 N.

The radial force on a gear can be calculated by the formula Fr = Ftan(α), where Fr is the radial force, Ft is the tangential force (in this case, the torque), and α is the pressure angle. The tangential force is equal to the torque divided by the pitch circle radius (i.e., Ft = T/r). Therefore, the radial force can be written as Fr = (T/r)tan(α).

To solve the problem, we need to find the pitch circle radius, which is equal to half the pitch circle diameter. So, r = 10 mm. We also know the torque (T = 1600 N.m) and the pressure angle (α = 18°). Plugging these values into the formula, we get:

Fr = (T/r)tan(α)

Fr = (1600 N.m / 10 mm)tan(18°)

Fr ≈ 5041 N

Therefore, the radial force on the gear is approximately 5041 N.

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The type of lens that spreads parallel light is a the type of lens that spreads parallel light is a Diverging lenses

When a lens in the path of a beam of parallel rays causes the rays to diverge after refraction, it is referred to as a diverging lens. It always creates a virtual image and has a thinner core than its boundaries. Concave lenses(diverging lenses) function as diverging lenses because they cause light rays to diverge as they pass through them. Therefore, when used with air as the operating medium, a concave lens can be thought of as a diverging lens. The concave lens has several applications, including in telescopes, cameras, lasers, glasses, binoculars, and other optical devices.

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The frequency heard was 530 hz

explain about frequency ?

The frequency of a repeated event is the number of occurrences per unit of time. It is also known as temporal frequency to distinguish it from spatial frequency and ordinary frequency to distinguish it from angular frequency. Frequency is measured in hertz (Hz), which is equal to one (event) per second. The period is the reciprocal of the frequency since it represents the temporal span of one cycle in a recurring occurrence.  For example, if a heart beats 120 times per minute (2 hertz), the period, T (the time interval between beats), is half a second (60 seconds divided by 120 beats).

for this case we know that the speed of the sound is given by

Vs=340 m/s

Vc=55m/s

vs=55m/s

Fo=530Hz

fs=(V-V0/v-vs)x530

=(340-55/340-55)x530

=530Hz

therefore the frequency heared  was found to be 530Hz.

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