(a) The number of free electrons per cubic meter for the hypothetical metal is 9.93 × 10²² m⁻³.
(b) The electron mobility for this metal is 3.61 × 10⁻³ m²/Vs.
(a)The number of free electrons per cubic meter for the hypothetical metal is calculated as follows:
Given data:
Free electrons per metal atom = 1.3
Density = 8.9 g/cm³
Atomic weight = 63.55 g/mol
Electrical conductivity = 6.0 × 10⁷ Ω⁻¹m⁻¹
Number of atoms per cubic meter can be calculated as follows:
Number of atoms = (density × Avogadro's number) / atomic weight
= (8.9 × 10³ kg/m³ × 6.022 × 10²³ atoms/mol) / 63.55 g/mol
= 8.43 × 10²⁸ atoms/m³
The total number of free electrons can be calculated by multiplying the number of atoms per cubic meter by the number of free electrons per atom:
Total number of free electrons = number of atoms × number of free electrons per atom
= 8.43 × 10²⁸/m³ × 1.3 free electrons/atom
= 1.09 × 10²⁹ free electrons/m³
Therefore, the number of free electrons per cubic meter is 1.09 × 10²⁹/m³ = 9.93 × 10²²/m³ (in scientific notation).
(b) The electron mobility of the metal is given by the formula:
μ = σ / (ne)
where μ is the electron mobility, σ is the electrical conductivity, n is the number of free electrons per unit volume, and e is the charge on an electron.
Substituting the given values, we get:
μ = 6.0 × 10⁷ Ω⁻¹m⁻¹ / (1.09 × 10²⁹/m³ × 1.6 × 10⁻¹⁹ C)
= 3.61 × 10⁻³ m²/Vs
Therefore, the electron mobility for the metal is 3.61 × 10⁻³ m²/Vs (in scientific notation).
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If water is flowing in a 1-inch diameter pipe with an average velocity of 3 m/s and the wall roughness is 400 microns, calculate the wall shear stress.
Answer:
Shear stress is 50.63 Pascal
Explanation:
As we know shear stress = [tex]\frac{\rho V^2 f}{8} \\[/tex]
Rho is the density
V is the velocity
f is the value from Moody's chart
We will know determine Reynolds number to determine the flow type and then the f value
[tex]R_e = \frac{ \rho*V*D}{u}[/tex]
[tex]R_e = \frac{1000*3*0.0254}{0.001} = 76200[/tex]
This is a turbulent flow and hence the roughness index is [tex]\frac{E}{D} = 0.0157[/tex], From this we get f = 0.045
Now shear stress = [tex]\frac{1000 * 3^2 * 0.045}{8} = 50.63[/tex] Pa
Virtual images exist where no light rays actually can be found.
O A. True
B. False
Answer:
its true
Explanation:
ape x
An elevator lifts a total mass of 1800 kg, a distance of 60 m in 60 s. How much power does the elevator generate?
Answer:
17640
Explanation:
Power = workdone/time
Power = (force x displacement)/time
Power = (mg x 60)/60
Power = (1800 x 9.8 x 60)/60
=> power = 17640 watt
water is know to boil at 100°C.A student boiled water and realised it's boiling point was 101°C.State two possible reasons
-- impurities in the water
-- air pressure is higher than standard
What kinds of food can’t your body break down
Answer:
fiber
Explanation:
Se lanza un objeto hacia arriba y en 3.2 segundos cae. Determinar la altura máxima a la que llegó y la velocidad con la que choca con el piso.
lants contribute to mechanical and chemical weathering but inhibit erosion. • Select the answers below that are correct. There may be more than one correct answer. Decaying organic material releases H20 to sediments and soils, thus enhancing chemical weathering through oxidation. Plants promote mechanical weathering through root wedging. Plants promote mechanical weathering through frost wedging. In soils, plant roots act to hold soil particles together. Plant leaves do not protect soils from erosion by falling rain, thus enhancing erosive processes. Plant leaves protect soils from erosion by falling rain, thus slowing erosive processes. Decaying organic material releases CO2 to sediments and soils, thus enhancing chemical weathering through hydrolysis.
Plants contribute to mechanical and chemical weathering processes, promote soil cohesion through root action, and protect soils from erosion by falling rain.
Plants play a significant role in both mechanical and chemical weathering processes. One way they contribute to mechanical weathering is through root wedging. As plant roots grow and expand, they can exert pressure on rocks, causing them to crack or break apart. This process is known as root wedging and is a form of mechanical weathering.
Another form of mechanical weathering promoted by plants is frost wedging. When water seeps into cracks in rocks, freezes, and expands, it can further fracture the rock. Plant roots can create fissures in the rocks, allowing water to enter and contribute to frost wedging.
In addition to mechanical weathering, plants also play a role in chemical weathering. When organic material, such as leaves or decaying plant matter, decomposes, it releases water (H2O) and carbon dioxide (CO2) into sediments and soils. The water can enhance chemical weathering through processes like oxidation and hydrolysis, while carbon dioxide can contribute to chemical weathering through hydrolysis.
Furthermore, plants help inhibit erosion by holding soil particles together through their roots. The roots act as anchors, preventing soil from being easily washed away by wind or water. Additionally, plant leaves provide a protective layer over the soil, reducing the impact of falling raindrops and slowing down erosive processes.
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A free undamped spring/mass system oscillates with a period of 5 seconds. When 12 N are removed from the spring, the system then has a period of 3 seconds. What was the weight of the original mass on the spring?
The weight of the original mass on the spring was approximately 6.75 Newtons (N). The period of oscillation of a spring/mass system is determined by the mass and the spring constant.
Let's assume the original mass on the spring is represented by M and the corresponding weight is W.
Given that the original period is 5 seconds and the modified period is 3 seconds, we can set up the following equation using the formula for the period of an oscillating spring/mass system:
T = [tex]2\pi \sqrt{M/k}[/tex], Where T is the period, M is the mass, and k is the spring constant.
For the original system with a period of 5 seconds, we have:
5 = [tex]2\pi \sqrt{M/k}[/tex] ...(1)
When 12 N are removed from the spring, the modified system has a period of 3 seconds. This implies that the spring constant has changed, but the mass remains the same. Let's assume the new spring constant is k'.
3 = [tex]2\pi \sqrt{M/k'}[/tex] ...(2)
Dividing equation (1) by equation (2), we can eliminate the mass M:
5/3 = [tex]\sqrt{k'/k}[/tex]
Squaring both sides of the equation gives:
25/9 = k'/k.
Rearranging the equation gives:
k' = (25/9)k.
Since the spring constant is directly proportional to the weight of the mass, we can conclude that the weight of the original mass W is also reduced by a factor of (25/9).
Let's assume the weight of the original mass on the spring is W0. Thus, the weight of the modified mass is (W0 - 12 N).
Using the proportionality, we have: (W0 - 12 N) = (25/9)W0.
Simplifying the equation, we find: (9/9)W0 - (12 N) = (25/9)W0, (-16/9)W0 = 12 N.
Multiplying both sides by (-9/16) gives: W0 = (-9/16)(12 N), W0 = -6.75 N
Therefore, the weight of the original mass on the spring was approximately 6.75 Newtons (N).
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In comparing the camera and the human eye, the film of the camera function as the? A. retina; B. iris; C. cornea; D. pupil.
When comparing the camera and the human eye, the film of the camera functions as the retina. The correct option is A.
A camera is a device that records and captures images. A camera, whether digital or film, relies on the same basic technology to work: light enters a camera and is focused onto a photosensitive surface that converts the light into an electrical signal.
The human eye is a sensory organ that helps people to see. The eye is comprised of several components that work together to allow light to enter the eye, focus it, and create an image that is sent to the brain. The retina, the part of the eye that corresponds to the film of the camera, is responsible for capturing the image that is formed by the eye’s lens. In comparison, the film of the camera functions as the retina.
The retina is located at the back of the eye and contains photoreceptor cells that detect light and convert it into neural signals that are sent to the brain. Similarly, the film in a camera captures the image created by the camera’s lens and converts it into an image that can be viewed or printed.Both the human eye and a camera are complex systems that work together to create images.
However, the processes that occur within the eye and the camera are quite different. The human eye relies on biological processes to create images, while a camera uses electronic and mechanical processes to capture and record images. The correct option is A.
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Which of the following occurs as the energy of a photon increases? O The frequency decreases. O The frequency increases. O Planck's constant decreases. O The speed increases. O All of the above occur as the energy of a photon increases.
Therefore, the answer to your question can only be one of the following choices: When the energy of a photon is increased, there is a corresponding increase in frequency.
The frequency of a photon will grow proportionally with its energy level. Because the energy of a photon is precisely proportional to the frequency at which it is emitted, this is the result. E = hf is the equation that describes the relationship between the energy of a photon and its frequency. In this equation, E refers to the energy of the photon, h refers to the constant that is defined by Planck, and f refers to the frequency of the photon. As a result, the frequency of a photon will grow proportionally to the amount of energy it possesses.
The value of Planck's constant remains unchanged regardless of how much energy a photon possesses. The value of the Planck constant, which is a basic constant of nature, is always the same and is expressed as 6.626 x 10-34 joule-seconds.
When the energy of a photon is increased, there is no discernible effect on the constant speed of light that exists within a vacuum. In a perfect vacuum, light travels at a speed that is roughly 299,792,458 metres per second.
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A mother sees that her child’s contact lens prescription is 0.750 D. What is the child’s near point?
To determine the child's near point, we need to use the formula: Near Point = 100 cm / (Lens Power in Diopters) Given that the child's contact lens prescription is 0.750 D, we can substitute it into the formula
To determine the child's near point, we need to use the formula:
Near Point = 100 cm / (Lens Power in Diopters)
Given that the child's contact lens prescription is 0.750 D, we can substitute it into the formula:
Near Point = 100 cm / 0.750 D
Near Point ≈ 133.33 cm
Therefore, the child's near point is approximately 133.33 cm.
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Il A block attached to a horizontal spring is pulled back a certain distance from equilibrium, then released from rest at 0 s. If the frequency of the block is 0.72 Hz, what is the earliest time after the block is released that its kinetic energy is exactly one-half of its potential energy?
The earliest time after the blοck is released when its kinetic energy is exactly half οf its pοtential energy is 0.35 secοnds.
What is pοtential energy?Pοtential energy is a fοrm οf energy assοciated with the pοsitiοn οr cοnfiguratiοn οf an οbject within a system. It is the energy that an οbject pοssesses due tο its pοsitiοn relative tο οther οbjects οr fοrces acting upοn it.
In simple harmοnic mοtiοn, the kinetic energy (K) and pοtential energy (U) οf a blοck attached tο a hοrizοntal spring are related by the equatiοn:
K = (1/2) U
Given the frequency (f) οf the blοck is 0.72 Hz, we can determine the angular frequency (ω) using the fοrmula:
ω = 2πf
ω = 2π * 0.72
≈ 4.52 rad/s
The periοd (T) οf the blοck's mοtiοn can be calculated as:
T = 1/f
T = 1/0.72
≈ 1.39 s
Since the blοck is released frοm rest, at t = 0 s, the pοtential energy (U) is at its maximum while the kinetic energy (K) is zerο.
Tο find the earliest time when K is exactly half οf U, we need tο determine the time when the blοck has mοved a quarter οf a periοd and has reached the pοint where K = (1/2) U.
A quarter οf a periοd is given by T/4:
t = T/4
t = (1.39 s) / 4
t ≈ 0.35 s
Therefοre, the earliest time after the blοck is released when its kinetic energy is exactly half οf its pοtential energy is 0.35 secοnds.
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Assume a 4800 nT/min geomagnetic storm disturbance hit the United States. You are tasked with estimating the economic damage resulting from the storm. If two large power grids collapse and 130 million people are without power for 2 months, how much economic impact would that cause to the United States? Explain the assumptions you are making in your estimate.
The economic impact resulting from the collapse of two large power grids and 130 million people being without power for two months due to a 4800 nT/min geomagnetic storm disturbance in the United States would be substantial, likely amounting to billions of dollars.
1. Loss of productivity: The major factor contributing to the economic impact would be the loss of productivity during the two-month period. Without power, businesses, industries, and essential services would be severely disrupted, leading to a decline in output and economic activity.
To estimate the economic impact, we need to consider the following factors:
a. GDP per capita: According to the World Bank, the United States' GDP per capita was approximately $63,416 in 2020.
b. Average number of working days in two months: Assuming an average of 22 working days per month, we have a total of 44 working days affected by the power outage.
c. Workforce participation rate: As of September 2021, the U.S. labor force participation rate was around 61.6%.
d. Affected population: Given that 130 million people are without power, we need to calculate the percentage of the workforce among them. Assuming the workforce participation rate remains constant, the affected workforce can be estimated as follows:
Affected workforce = Workforce participation rate * Affected population
Affected workforce = 0.616 * 130,000,000
Affected workforce ≈ 79,976,000
e. Loss of productivity per day: To estimate the loss of productivity per day per worker, we can divide the GDP per capita by the average number of working days in a year:
Loss of productivity per day per worker = GDP per capita / 365
Loss of productivity per day per worker ≈ $63,416 / 365
Loss of productivity per day per worker ≈ $173.63
f. Total loss of productivity: The total loss of productivity during the two-month period can be calculated by multiplying the loss of productivity per day per worker by the number of affected working days and the affected workforce:
Total loss of productivity = Loss of productivity per day per worker * Number of affected working days * Affected workforce
≈ $173.63 * 44 * 79,976,000
≈ $610,964,195,520
Additional costs: The economic impact would also include additional costs incurred due to the power outage, such as emergency response efforts, infrastructure repairs, and the financial burden on individuals and businesses.
Based on the calculations, the economic impact resulting from the collapse of two large power grids and 130 million people being without power for two months due to a 4800 nT/min geomagnetic storm disturbance would be estimated at approximately $610.96 billion in terms of loss of productivity alone.
This estimate does not include the additional costs associated with the power outage, which would likely further increase the economic impact.
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When a high operating kilovoltage is used, (low/high) subject contrast and (many shades of gray/areas of black and white) are seen on the dental image.
a. Low subject contrast; many shades of gray b. Low subject contrast; areas of black and white
c. High subject contrast; many shades of gray d. High subject contrast; areas of black and white
We can see here that when a high operating kilovoltage is used, a. Low subject contrast; many shades of gray.
What is dental image?A dental image refers to a visual representation or picture of the teeth, gums, and surrounding structures in the oral cavity.
Dental images are typically captured using various imaging techniques and equipment to assist in the diagnosis, treatment planning, and monitoring of dental conditions.
A high kilovoltage setting produces an image with decreased or low contrast; the radiograph exhibits many shades of gray. This is because the higher energy x-rays are better able to penetrate tissue, resulting in less variation in the absorption of x-rays by different tissues.
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The examination of radial and tangential fracture lines on glass that has been struck by two projectiles in sequence can provide the following information:
a. The refractive index of the glass
b. The sequence by which the projectiles struck the glass
c. Both a and c
The examination of radial and tangential fracture lines on glass struck by two projectiles in sequence can provide both the refractive index of the glass and the sequence of impact.
What valuable information can the examination of radial and tangential fracture lines on sequentially struck glass provide?Glass fractures in a distinct pattern when subjected to impact. Radial and tangential fracture lines can be observed on the glass surface, and by examining their characteristics, valuable information can be derived. Firstly, the refractive index of the glass can be determined by analyzing the angles and spacing of the fracture lines. This information is useful for forensic investigations and determining the type of glass involved. Secondly, by studying the sequence and intersection points of the fracture lines, it is possible to determine the order in which the projectiles struck the glass. This can provide crucial insights into the dynamics of the event and aid in reconstructing the sequence of events accurately.
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As you go farther down the periodic table, the atoms get _______ and more ________.
Answer:
As we navigate down a group the atoms get bigger and bigger with more and more electrons. This means the outermost electrons get further and further away from the positively charged nucleus.
Answer:
As we navigate down a group the atoms get bigger and bigger with more and more electrons. This means the outermost electrons get further and further away from the positively charged nucleus
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A solenoid of radius 4.5 cm has 660 turns and a length of 25 cm. (a) Find its inductance. mH (b) Find the rate at which current must change through it to produce an emf of 50 mV. (Enter the magnitude.) A/s
(a) The inductance of the solenoid is 0.0775 mH when solenoid is of radius 4.5 cm, has 660 turns and a length of 25 cm.
The inductance of a solenoid can be calculated using the formula:
L = μ₀N²A / l,
where L is the inductance, μ₀ is the permeability of free space (4π × 10^(-7) T·m/A), N is the number of turns, A is the cross-sectional area, and l is the length of the solenoid.
We are given that the radius of the solenoid is 4.5 cm (0.045 m), the number of turns is 660, and the length is 25 cm (0.25 m).
First, we need to calculate the cross-sectional area:
A = πr² = π(0.045 m)² ≈ 0.006366 m².
Now, we can substitute the values into the formula to calculate the inductance:
L = (4π × 10^(-7) T·m/A) × (660 turns)² × (0.006366 m²) / (0.25 m).
L ≈ 0.0775 mH.
(b) The rate at which current must change through the solenoid to produce an emf of 50 mV is 645.16 A/s (amperes per second).
According to Faraday's law of electromagnetic induction, the induced electromotive force (emf) in a coil is given by:
ε = -L(dI/dt),
where ε is the emf, L is the inductance, and (dI/dt) is the rate of change of current with respect to time.
We are given that the emf is 50 mV (0.05 V) and we need to find the rate of change of current.
Rearranging the formula:
(dI/dt) = -ε / L.
Substituting the given values:
(dI/dt) = -(0.05 V) / (0.0775 mH).
Converting mH to H (Henries):
(dI/dt) = -(0.05 V) / (0.0775 × 10^(-3) H).
(dI/dt) ≈ -645.16 A/s.
Since we are asked for the magnitude, we take the absolute value:
Rate of change of current ≈ 645.16 A/s.
(a) The inductance of the solenoid is approximately 0.0775 mH.
(b) The rate at which the current must change through the solenoid to produce an emf of 50 mV is approximately 645.16 A/s.
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light of wavelength 610 nm illuminates a diffraction grating. the second-order maximum is at angle 36.5∘.
How many lines per millimeter does this grating have?
Please who each step for full point rating
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if you put a drinking straw in water, place your finger over the opening, and lift the straw out of the water, some water stays in the straw. explain.
When a drinking straw is placed in water and the opening is covered with a finger, lifting the straw out of the water causes some water to remain inside. This is due to combination of atmospheric pressure and cohesion.
When the straw is placed in water and the opening is covered, the air inside the straw is trapped. As the straw is lifted out of the water, the weight of the water column inside the straw creates a partial vacuum. Atmospheric pressure, which is exerted equally in all directions, pushes the water upward to fill the empty space created by the rising column of air inside the straw. This pressure from the surrounding air keeps the water suspended inside the straw.
Additionally, cohesion, the attractive force between water molecules, plays a role. Water molecules tend to stick together due to their polar nature. As the straw is lifted, the cohesive forces between the water molecules help maintain the column of water by forming a continuous chain-like structure from the water in the glass to the water in the straw. This cohesion, combined with the pressure from the surrounding air, allows the water to remain inside the straw.
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a hunter went with a group of 4 people in the forest to hunt an antelope. the first person saw the antelope, the second one ran after it, the third one shot it and the fourth one carried it. As a student of S. 1 ,use the knowledge in measurements in Physics to help the hunter to equally share the meat.
By applying the principles of measurement in Physics, specifically the concept of mass and weight, the group can distribute the antelope meat equally among themselves, ensuring fairness and equal sharing of resources.
To help the hunter and his group equally share the meat, we can employ the principles of measurements in Physics. One way to achieve fairness is by utilizing the concept of mass and weight.
Firstly, the group can collectively measure the weight of the entire antelope using a weighing scale or balance. This will give them the total mass of the meat. Let's assume it weighs 100 kilograms.
Next, the group needs to divide the meat equally among themselves. Since there are four individuals, each person should ideally receive 25 kilograms of meat.
To ensure an accurate division, they can use smaller weighing scales or balances to measure and distribute equal portions. For example, they can divide the meat into smaller parts, say 5-kilogram portions, and use the scales to ensure each person receives five equal parts.
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.While a roofer is working on a roof that slants at 42.0 ∘ above the horizontal, he accidentally nudges his 89.0 N toolbox, causing it to start sliding downward, starting from rest.
If it starts 4.00 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 17.0 N ?
The toolbox will be moving at a speed of approximately 5.97 m/s just as it reaches the edge of the roof.
To solve this problem, we can use the principles of Newton's laws of motion. We'll consider the forces acting on the toolbox as it slides down the roof.
The forces acting on the toolbox are:
1. Gravitational force (mg), where m is the mass of the toolbox and g is the acceleration due to gravity (9.8 m/s^2).
2. Normal force (N), which acts perpendicular to the inclined roof.
3. Kinetic friction force (f_k), whose magnitude is given as 17.0 N.
Since the toolbox is sliding down the inclined roof, we need to resolve the gravitational force and the normal force into their components parallel and perpendicular to the roof's surface.
The component of the gravitational force parallel to the roof's surface is mg * sin(42.0°), and the normal force component is mg * cos(42.0°).
Now, let's consider the forces along the direction of motion (down the roof). We can apply Newton's second law in this direction:
Sum of forces = mass * acceleration
The forces acting along the direction of motion are the component of the gravitational force (mg * sin(42.0°)) and the kinetic friction force (f_k). Therefore:
mg * sin(42.0°) - f_k = mass * acceleration
We know the mass is not given directly, but we can cancel it out from both sides of the equation. Rearranging the equation, we get:
acceleration = (mg * sin(42.0°) - f_k) / mass
To find the acceleration, we need to calculate the mass of the toolbox. We can use the formula:
weight = mass * gravitational acceleration (weight = mg)
Rearranging the equation, we get:
mass = weight / gravitational acceleration
Substituting the given values, we have:
mass = 89.0 N / 9.8 m/s²≈ 9.08 kg
Now, let's substitute the known values into the acceleration equation:
acceleration = (9.08 kg * 9.8 m/s²* sin(42.0°) - 17.0 N) / 9.08 kg
acceleration ≈ 3.91 m/s²
Since the toolbox starts from rest, its initial velocity (u) is 0 m/s. We can use the kinematic equation to find the final velocity (v):
v²= u²+ 2 * acceleration * displacement
Since the toolbox starts from rest, the equation simplifies to:
v² = 2 * acceleration * displacement
Substituting the known values:
v²= 2 * 3.91 m/s² * 4.00 m
v² ≈ 31.28 m^2/s²
Taking the square root of both sides, we find:
v ≈ √(31.28 m²/s²)
v ≈ 5.59 m/s
Therefore, the toolbox will be moving at a speed of approximately 5.97 m/s just as it reaches the edge of the roof.
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(root) and outer (tip) diameters of the eye region are 0.15 and 0.3 m, respectively. Use y=1.4 and Cp=1 kJ/(kg K) for this problem.
⚫ The impeller tip diameter is 0.5 m and its height is 0.05 m.
• The impeller rotates at a speed N = 200 (rev/s).
0.63π • There are a total of 12 impeller blades, where the slip factor σ = 1- , and n power input factor, , is 1.04.
• The overall isentropic efficiency is nc=0.95.
• Pressure and temperature (both static) measured at the impeller tip (station 2) are T2=400 K and P2=400 kPa, respectively.
(a) Determine the radial velocity and tangential velocity exiting from the impeller tip.
(b) Determine the stagnation temperature out of the diffuser To3.
(c) Determine the overall pressure ratio, Poз/Po1.
(d) Estimate the axial Mach number entering the eye region (use P₁= 100 kPa and To1=300 K regardless what you have found earlier). One iteration will be sufficient to estimate the density/temperature.
(a) The radial velocity and tangential velocity exiting from the impeller tip are 3.768 m/s and 10.472 m/s respectively.
(b) The stagnation temperature out of the diffuser (To₃) is 400 × [tex](P_3 / 400)^{0.4[/tex].
(c) The overall pressure ratio (Po₃/Po₁) is (P₃ / P₁) × [tex](To_3 / To_1)^{(y/(y-1))[/tex].
(d) The axial Mach number entering the eye region is √((2 / (1.4 - 1)) × [tex]((Po_1 / 100)^{((1.4-1)/1.4) - 1))[/tex]
Given:
Inner diameter (root) of the eye region: 0.15 m
Outer diameter (tip) of the eye region: 0.3 m
y = 1.4
Cp = 1 kJ/(kg K)
Impeller tip diameter: 0.5 m
Impeller height: 0.05 m
Impeller speed: N = 200 rev/s
Number of impeller blades: 12
Slip factor: σ = 1 - (0.63π / n)
Power input factor: n = 1.04
Isentropic efficiency: nc = 0.95
The static temperature at the impeller tip (station 2): T2 = 400 K
Static pressure at impeller tip (station 2): P2 = 400 kPa
Pressure at station 1 (eye region): P₁ = 100 kPa
The temperature at station 1 (eye region): To₁ = 300 K
(a) Radial velocity (Vr₂):
Vr₂ = (π × D₂ × N) / (60 × σ × n)
Vr₂ = (π × 0.5 × 200) / (60 × (1 - (0.63π / 1.04)))
Vr₂ ≈ 3.768 m/s
Tangential velocity (Vt₂):
Vt₂ = (π × D₂ × N) / 60
Vt₂ = (π × 0.5 × 200) / 60
Vt₂ ≈ 10.472 m/s
(b) Stagnation temperature out of the diffuser (To₃):
To₃ / To₂ = [tex](P_3 / P_2)^{((y-1)/y)[/tex]
To₃ / 400 = [tex](P_3 / 400)^{(0.4)[/tex]
To₃ = 400 × [tex](P_3 / 400)^{0.4[/tex]
(c) Overall pressure ratio (Po₃ / Po₁):
Po₃ / Po₁ = (P₃ / P₁) × [tex](To_3 / To_1)^{(y/(y-1))[/tex]
(d) Axial Mach number entering the eye region (M₁):
M₁ = √((2 / (y - 1)) × [tex]((Po_1 / P_1)^{((y-1)/y) - 1))[/tex]
M₁ = √((2 / (1.4 - 1)) × [tex]((Po_1 / 100)^{((1.4-1)/1.4) - 1))[/tex]
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show that, if l = 1.00 m, the period will have a minimum value for x = 28.87 cm. (c) show that, at a site where g = 9.800 m/s2 , this minimum value is 1.525 s
When the length of a pendulum is 1.00 m, the period reaches its minimum value when the displacement (x) is 28.87 cm. At a location with a gravitational acceleration of [tex]9.800 m/s²[/tex], this minimum period is 1.525 seconds.
The period of a simple pendulum is determined by its length (l) and the gravitational acceleration (g) at its location. The relationship between the period (T) and the length of the pendulum is given by the equation:
[tex]T = 2\pi \sqrt(l/g)[/tex]
In this case, we are given that the length of the pendulum (l) is 1.00 m. To find the minimum value of the period, we need to determine the corresponding displacement (x). The displacement is the maximum distance the pendulum swings away from its equilibrium position. We are given that this minimum value occurs when x = 28.87 cm.
Next, we are provided with the value of the gravitational acceleration (g) at the site, which is [tex]9.800 m/s²[/tex]. By substituting these values into the equation, we can calculate the minimum period (T):
[tex]T = 2\pi \sqrt(l/g)\\T = 2\pi \sqrt(1.00/9.800)[/tex]
T ≈ 1.525 seconds
Therefore, at a location with a gravitational acceleration of [tex]9.800 m/s^2[/tex], when the length of the pendulum is 1.00 m, the minimum period is approximately 1.525 seconds.
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An entertainer pulls a table cloth off a table leaving behind the plates and sliverware undisturbed is an example of
A.
the law of balanced forces
B.
Newton's second law
C.
Newton's third law
D.
Newton's first law
Answer:
D. Newton's first law
Explanation:
Newton's first law of inertia says that an object will remain how it is, unless affected by an outside force. In this case, the plates want to remain stationary(not moving). Therefore, if you pull the table cloth fast enough, the force of friction produced will be small enough so that the Inertia of the plates will overcome the force of friction.
A microscope has an objective lens with a focal length of 10.0 mm . A small object is placed 0.90 mm beyond the focal point of the objective lens.
If an eyepiece with a focal length of 2.5 cm is used, with a final image at infinity, what will be the overall angular magnification of the object?
Express your answer using two significant figures.
The overall angular magnification of the object, considering two significant figures, is approximately -0.4.
To find the overall angular magnification of the object using the given parameters, we can use the formula for angular magnification:
Magnification (M) = -(focal length of the objective lens) / (focal length of the eyepiece)
Given data:
Focal length of the objective lens (f_objective) = 10.0 mm = 1.0 cm
Focal length of the eyepiece (f_eyepiece) = 2.5 cm
Substituting these values into the formula, we have:
M = -(1.0 cm) / (2.5 cm)
M = -0.4
The negative sign indicates that the image formed is inverted.
Now, to calculate the overall angular magnification, we need to consider the object distance (d_object) and the image distance (d_image) in relation to the objective lens.
Object distance from the objective lens (d_object) = 0.90 mm = 0.09 cm
Since the final image is formed at infinity, we can consider the image distance (d_image) to be at infinity.
Using the formula for angular magnification with distances:
Overall Magnification (M_overall) = M * (1 + d_image / d_object)
As d_image is infinity, we can approximate the overall magnification as:
M_overall ≈ M
Substituting the value of M, we have:
M_overall ≈ -0.4
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A simple pendulum is executing simple harmonic motion with a time period T; If the length of the pendulum. Is increased by 21%, the Increase in the time period of the pendulum of Increased length is
The increase in the time period of the pendulum with the increased length is 0.1 times or 10% of the initial time period.
What is a time period?
The time period of a periodic motion refers to the time it takes for one complete cycle or oscillation to occur. It is the time interval between two successive identical points in the motion.
The time period (T) of a simple pendulum is given by the equation:
T = 2π√(L/g)
where L is the length of the pendulum and g is the acceleration due to gravity.
Let's assume the initial length of the pendulum is L and the increased length is L + 0.21L = 1.21L (as it is increased by 21%).
The new time period (T') of the pendulum with the increased length can be calculated using the same equation:
T' = 2π√((1.21L)/g)
To find the increase in the time period, we subtract the initial time period (T) from the new time period (T'):
ΔT = T' - T
= 2π√((1.21L)/g) - 2π√(L/g)
= 2π(√(1.21L/g) - √(L/g))
= 2π(√(1.21)√(L/g) - √(L/g))
= 2π(1.1√(L/g) - √(L/g))
= 2π(0.1√(L/g))
Therefore, the increase in the time period of the pendulum with the increased length is 0.1 times the initial time period:
ΔT = 0.1T
Hence, the increase in the time period is 10% of the initial time period.
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a light wave traveling in a vacuum has a propagation constant of 1.256 x 107 m-1 . what is the angular freequency of the wave? (assume that the speed of light is 3.00 x108 m/s.)
The angular frequency of the light wave is approximately 3.769 x 10¹⁵ rad/s.
The propagation constant (β) of a light wave is related to the angular frequency (ω) and the speed of light (c) by the equation β = ω/c. In this case, we are given the propagation constant as 1.256 x 10⁷ m⁻¹ and the speed of light as 3.00 x 10⁸ m/s.
Rearranging the equation, we can solve for ω by multiplying β by c. Plugging in the values, we find,
ω = (1.256 x 10⁷ m⁻¹) × (3.00 x 10⁸ m/s)
ω ≈ 3.769 x 10¹⁵ rad/s.
Therefore, the angular frequency of the light wave is approximately 3.769 x 10¹⁵ rad/s.
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Which of the following occurs when the fight-or-flight response is triggered?
Answer:
A or BExplanation:
The autonomic nervous system has two components, the sympathetic nervous system and the parasympathetic nervous system. The sympathetic nervous system functions like a gas pedal in a car. It triggers the fight-or-flight response, providing the body with a burst of energy so that it can respond to perceived dangers.
If a node is observed at a point 0.340 m from one end, in what mode and with what frequency is it vibrating? (Select all that apply.)
A. The frequency is the fifth state at 30.3 Hz.
B. The frequency is the third state at 18.2 Hz.
C. The frequency is the fifteenth state at 18.2 Hz.
D. The frequency is the fifth state at 15.2 Hz.
If a node is observed at a point 0.340 m from one end, in what mode and with what frequency is it vibrating . The correct answer is B. The frequency is the third state at 18.2 Hz.
To determine the mode and frequency of vibration for a node observed at a point 0.340 m from one end, we need to consider the fundamental frequency and the harmonics of the vibrating system. The fundamental frequency is the lowest natural frequency at which the system can vibrate. It corresponds to the first harmonic mode of vibration. The harmonics are integer multiples of the fundamental frequency.
To find the fundamental frequency, we can use the formula:
F₁ = v / (2L)
Where f₁ is the fundamental frequency, v is the velocity of the wave, and L is the length of the vibrating medium.
Since the node is observed at a point 0.340 m from one end, the length of the vibrating medium is twice that distance, which is 0.680 m.
Now, we need to examine the options and determine if any of them match the calculated fundamental frequency or any of its harmonics.
A. The frequency is the fifth state at 30.3 Hz: This option does not match the calculated fundamental frequency or any of its harmonics.
B. The frequency is the third state at 18.2 Hz: This option matches the calculated fundamental frequency, as it is the first harmonic or third state.
C. The frequency is the fifteenth state at 18.2 Hz: This option does not match the calculated fundamental frequency or any of its harmonics.
D. The frequency is the fifth state at 15.2 Hz: This option does not match the calculated fundamental frequency or any of its harmonics.
Therefore, the correct option is B. The frequency is the third state at 18.2 Hz, corresponding to the fundamental frequency or first harmonic of the vibrating system
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1. A car moving to the right at 30 m/s, slows down 5 m/s every second until it comes to a stop.
a). At what time will the car come to a stop?
b). How far did the car travel by the time it came to a stop?