In a solid long conducting cylinder carrying a net negative charge, the electric field is zero inside the conductor, and radially inward outside the conductor. Therefore, the option D is correct.
The electric field is zero inside a solid, long conducting cylinder. This is because in a conductor in electrostatic equilibrium, the charges redistribute themselves such that the internal electric field is zero.
The electric field is radially inward outside the conductor because an inward electric field outside the conductor is created when a negatively charged cylinder draws positively charged particles to its surface.
Hence, with a solid, long conducting cylinder carrying a net negative charge, option D correctly describes the electric field.
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The experimenter had observed that some colors of birthday balloons seem to be harder to inflate than others. She ran this experiment to determine whether balloons of different colors are similar in terms of the time taken for inflation to a diameter of 7 inches. Four colors were selected from a single manufacturer. An assistant blew up the balloons and the experimenter recorded the times (to the nearest 1/10 second) with a stop watch. Questions for all the following cases: Please identify: Independent variable and number of level? Dependent variable? Study design (i.e., between or within-subject design)? Confounding variable (if any)? Violation of Validity (if measureable)?
Case: The effect of balloon color on inflation time.
Independent variable: Balloon color (categorical) with four levels (e.g., red, blue, green, yellow).
Dependent variable: Time taken for inflation to a diameter of 7 inches (continuous, measured in seconds).
Study design: Within-subject design (the same group of participants inflating balloons of different colors).
Confounding variable: Possible confounding variables could be the size or material of the balloons, as these factors might affect the inflation time. To control for this, it would be important to ensure that all balloons used in the experiment are of the same size and material.
Violation of Validity: A violation of validity could occur if the measurement of inflation time is not accurate or consistent (e.g., if the stopwatch used is unreliable or if the experimenter's recording of times is inconsistent). Ensuring proper measurement procedures and equipment would help mitigate this violation.
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When throwing a ball, your hand releases it at a height of 1.1 m above the ground with velocity 6.4 m/s in direction 53° above the horizontal. (a) How high above the ground (not your hand) does the ball go? ___ m (b) At the highest point, how far is the ball horizontally from the point of release? ___m
By using the equations of projectile motion, the ball reaches a height of approximately 2.355 meters above the ground.
To solve this problem, we can use the equations of projectile motion.
(a) To determine the maximum height reached by the ball, we can analyze the vertical motion. We'll use the following kinematic equation:
Vertical displacement (Δy) = (Initial vertical velocity (v₀y) x Time of flight (t)) - (0.5 x Acceleration due to gravity (g) x Time of flight (t)²)
The initial vertical velocity can be found using the given initial velocity and the angle of projection:
v₀y = Initial velocity (v₀) x sin(angle)
v₀y = 6.4 m/s x sin(53°)
Now we can find the time of flight using the equation:
Time of flight (t) = (2 x v₀y) / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values, we get:
t = (2 x (6.4 m/s x sin(53°))) / 9.8 m/s²
Next, we substitute the time of flight back into the first equation to find the vertical displacement:
Δy = (6.4 m/s x sin(53°)) x ((2 x (6.4 m/s x sin(53°))) / 9.8 m/s²) - (0.5 x 9.8 m/s²) x ((2 x (6.4 m/s x sin(53°))) / 9.8 m/s²)²
Simplifying the equation, we find:
Δy ≈ 2.355 m
Therefore, the ball reaches a height of approximately 2.355 meters above the ground.
(b) At the highest point of the ball's trajectory, its vertical velocity becomes zero. At this point, only the horizontal velocity component is active. The horizontal distance traveled can be determined using the equation:
Horizontal distance = Horizontal velocity x Time of flight
The horizontal velocity can be found using the given initial velocity and the angle of projection:
Horizontal velocity = Initial velocity (v₀) x cos(angle)
Substituting the values, we get:
Horizontal distance = 6.4 m/s x cos(53°) x ((2 x (6.4 m/s x sin(53°))) / 9.8 m/s²)
Simplifying the equation, we find:
Horizontal distance ≈ 6.615 m
Therefore, the ball is approximately 6.615 meters horizontally from the point of release at its highest point.
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ACTIVITY 4
Applying the equation learned, answer the following problems:
1. A bowling ball whose mass is 4.0 kg is rolling at a rate of 2.5 m/s. What is its momentum? p = m/s. What Is Its Momentum?
Given:
Find:
Formula:
Solution:
2. A skateboard is rolling at a velocity of 3.0 m/s with a momentum of 6.0 kg-m/s. What is its mass?
Given:
Find:
Formula:
Solution:
3. A pitcher throws a baseball with a mass of 0.5 kg and a momentum of 10 kg-m/s. What is its velocity?
Given:
Find:
Formula:
Solution:
Subject Is Science
Good Perfect Complete=Brainlist
Copy Wrong Incomplete=Report
Good Luck Answer Brainly Users:-)
Answer:
1) 10 kg-m/s
2) 2 kg
3) 20 m/s
Explanation:
The momentum of an object can be calculated using the equation:
[tex]\large\boxed{p=mv}[/tex]
where:
p is momentum (measured in kilogram meters per second).m is mass (measured in kilograms).v is the velocity (measured in meters per second).[tex]\hrulefill[/tex]
Question 1For this question we need to find the momentum of a bowling ball whose mass is 4.0 kg is rolling at a rate of 2.5 m/s.
Given values:
m = 4.0 kgv = 2.5 m/sSubstitute the given values into the momentum formula and solve for p:
[tex]p=4.0\;\text{kg} \cdot 2.5\;\text{m/s}[/tex]
[tex]p=10\;\text{kg m/s}[/tex]
Therefore, the momentum of the bowling ball is 10 kg-m/s.
[tex]\hrulefill[/tex]
Question 2For this question we need to find the mass of a skateboard rolling at a velocity of 3.0 m/s with a momentum of 6.0 kg-m/s.
Given values:
p = 6.0 kg-m/sv = 3.0 m/sAs we want to find mass, rearrange the momentum formula to isolate m:
[tex]\large\boxed{m=\dfrac{p}{v}}[/tex]
Substitute the given values into the formula and solve for m:
[tex]m=\dfrac{6.0\; \text{kg m/s}}{3.0\; \text{m/s}}[/tex]
[tex]m=2\;\text{kg}[/tex]
Therefore, the mass of the skateboard is 2 kg.
[tex]\hrulefill[/tex]
Question 3For this question we need to find the velocity of a baseball with a mass of 0.5 kg and a momentum of 10 kg-m/s.
Given values:
p = 10 kg-m/sm = 0.5 kgAs we want to find velocity, rearrange the momentum formula to isolate v:
[tex]\large\boxed{v=\dfrac{p}{m}}[/tex]
Substitute the given values into the formula and solve for v:
[tex]v=\dfrac{10\; \text{kg m/s}}{0.5\; \text{kg}}[/tex]
[tex]v=20\;\text{m/s}[/tex]
Therefore, the velocity of the baseball is 20 m/s.
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You go into a room that is lit by green light only. You see a banana in the room.
a) What color does the banana look like in that room? Explain.
b) You see a shirt in that room. While in the room, the shirt appears to be green. Does the shirt reflect or absorb green light? Explain.
c. Can you tell if the shirt reflects or absorbs red light? What about blue light?
d. If you move the shirt out into a normally lit room (a room that has red, blue, and green light), list at least two possible colors the shirt could be. (Hint: red+green = yellow; red+blue = magenta; green+blue = cyan). Explain.
e. If you move the shirt out into a normally lit room, list at least two possible colors that the shirt could not be. Explain.
The banana would seem black or extremely dark in a room with just green lighting. This is so that items may be illuminated by green light sources, but a banana doesn't reflect green light.
If a shirt appears green in a room that is lighted by green light, it reflects green light. The shirt absorbs other colors of light while reflecting the green light that strikes it. The belief that the shirt is green in the room is caused by this green light's selective reflection.
In the room because there isn't enough red light to see how it interacts with the garment in a green-lit room, it is challenging to tell whether the shirt absorbs or reflects red light.
The shirt might appear in a variety of colors when placed in a room that is regularly lighted by red, blue, and green light.
Red and green light combine to form yellow, therefore if the garment reflects both colors, it will seem yellow.
Since magenta is created by combining red and blue light, if the garment reflects both red and blue light, it will seem that color.
There are a minimum of two colors that the shirt cannot be when it is placed in a room that is illuminated normally:
The fact that the shirt reflected green light in the green-lit space proves that it does not absorb green light, which rules out the possibility of a pure green appearance.
Because the shirt failed to reflect blue light in the green-lit space, indicating that it doesn't reflect or absorb blue light, the garment cannot seem pure blue.
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The temperature, u(x,t), of a metal bar of length 1 = 2.5 at a distance x from one end and at time t is modelled by the partial differential equation: Du/dt = a a2u/ax2
for 0 < x < 2.5 and t > 0. The metal has a thermal diffusivity a = 1.5 and the two ends of the bar are kept at a temperature, u(0,t) = u(2.5, t) = 0 and the initial temperature distribution is given by: u(x,0) = 2sin(╥x/l) (a) State the forward difference approximation for du/ât and du/ox and the central difference approximation, a2u/ax2
(b) Deduce the numerical scheme for approximating u(x, t) from the values of u(x, t) at the previous time step [2 marks] (c) Confirm that the choice of &x = 0.5 and 8t = 0.05 will yield a stable solution. (1 Mark) (d) Use the explicit difference scheme with 8x = 0.5 and 8t = 0.05 to approximate = u(x,t) at times t = 8t and t = 28t. [5 Marks)
(a) Forward difference approximation:
(∂u/∂t)ᵢⱼ ≈ (uᵢⱼ₊₁ - uᵢⱼ)/Δt
(∂u/∂x)ᵢⱼ ≈ (uᵢ₊₁ⱼ - uᵢⱼ)/Δx
Central difference approximation: (∂²u/∂x²)ᵢⱼ ≈ (uᵢ₊₁ⱼ - 2uᵢⱼ + uᵢ₋₁ⱼ)/Δx²
(b) Numerical scheme: uᵢⱼ₊₁ = uᵢⱼ + (a * Δt/Δx²) * (uᵢ₊₁ⱼ - 2uᵢⱼ + uᵢ₋₁ⱼ)
(c) For stability, the coefficient a * Δt/Δx² should be less than or equal to 0.5.
(d) To approximate u(x, t) at times t = 8t and t = 28t, we need to apply the numerical scheme iteratively using the given values of Δx = 0.5 and Δt = 0.05.
(a) The forward difference approximation for du/dt can be obtained by using the forward difference operator:
(∂u/∂t)ᵢⱼ ≈ (uᵢⱼ₊₁ - uᵢⱼ)/Δt
where (∂u/∂t)ᵢⱼ represents the approximation of du/dt at grid point (i, j), uᵢⱼ₊₁ is the temperature at the next time step, and Δt is the time step size.
Similarly, the forward difference approximation for du/dx can be obtained by using the forward difference operator:
(∂u/∂x)ᵢⱼ ≈ (uᵢ₊₁ⱼ - uᵢⱼ)/Δx
where (∂u/∂x)ᵢⱼ represents the approximation of du/dx at grid point (i, j), uᵢ₊₁ⱼ is the temperature at the next grid point in the x-direction, and Δx is the grid spacing in the x-direction.
The central difference approximation for a²u/∂x² can be obtained by using the central difference operator:
(∂²u/∂x²)ᵢⱼ ≈ (uᵢ₊₁ⱼ - 2uᵢⱼ + uᵢ₋₁ⱼ)/Δx²
where (∂²u/∂x²)ᵢⱼ represents the approximation of a²u/∂x² at grid point (i, j), uᵢ₊₁ⱼ and uᵢ₋₁ⱼ are the temperatures at the neighboring grid points in the x-direction, and Δx is the grid spacing in the x-direction.
(b) The numerical scheme for approximating u(x, t) can be obtained by substituting the forward difference approximations into the given partial differential equation:
Du/dt = a * a²u/∂x²
Using the forward difference approximations, we have:
(uᵢⱼ₊₁ - uᵢⱼ)/Δt = a * (uᵢ₊₁ⱼ - 2uᵢⱼ + uᵢ₋₁ⱼ)/Δx²
Simplifying the equation, we can rearrange it to solve for uᵢⱼ₊₁:
uᵢⱼ₊₁ = uᵢⱼ + (a * Δt/Δx²) * (uᵢ₊₁ⱼ - 2uᵢⱼ + uᵢ₋₁ⱼ)
This equation represents the numerical scheme for approximating u(x, t) at the next time step using the values of u(x, t) at the previous time step.
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Select all the statements that apply. 1. Young's modulus is a characteristic of a solid 2. Shear modulus is a characteristic of a solid 3. Shear stress results from a force exerted perpendicular to an area. 4. The bulk modulus is a characteristic of a liquid 5. Tensile stress results from a force exerted parallel to an area
A solid's Young's modulus is one of its properties. The ratio of the applied stress to the resulting strain serves as a metric for determining the stiffness of a solid material.
It is sometimes referred to as the elastic modulus and is normally expressed in pascals (Pa) units. A solid's shear modulus is another quality. It is a measurement of a solid material's stiffness in response to shear stress and is established as the relationship between the applied shear stress and the resulting strain.
Pascals (Pa) are the most common units used to measure it. When a force is applied perpendicular to an object, shear stress is the result. It is a form of tension brought on by a force.
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Which of the following statements about energy during a phase change is correct?
1. The average kinetic energy of the sample does not change during a phase change.
2.The average kinetic energy of the sample does decreases during a phase change.
3. The average potential energy of the sample does not change during a phase change.
4. The average kinetic energy of the sample increases during a phase change.
The correct statement about energy during a phase change is given by "1. The average kinetic energy of the sample does not change during a phase change."
During a phase change, such as the transition from solid to liquid or liquid to gas, the average kinetic energy of the particles remains constant.
Although energy is being transferred and absorbed or released during the phase change, this energy is primarily used to overcome intermolecular forces and change the arrangement of particles rather than increase their average kinetic energy.
The average potential energy of the sample can change during a phase change as the arrangement of particles and their interactions may vary.
Therefore, the average kinetic energy of the sample does not change during a phase change.
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1.
Write the original Faraday's law. Why did it have to be modified by
Lenz?
Faraday's law of electromagnetic induction states that the electromotive force induced in a circuit is directly proportional to the rate of change of magnetic flux through the circuit.
While Faraday's law successfully explained the phenomenon of electromagnetic induction, it did not address the direction of the induced current or emf. To fill this gap, Heinrich Lenz proposed Lenz's law, which states that the direction of the induced current is such that it opposes the change in magnetic flux that produced it.
Lenz's law is based on the principle of conservation of energy. When a magnetic field interacts with a conductor, causing a change in magnetic flux, an emf is induced to create an opposing current. This opposing current generates a magnetic field that opposes the change in the original magnetic field, thus conserving energy.
Lenz's modification of Faraday's law ensures that energy is conserved in electromagnetic processes.
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A nonconducting ring of radius 12 cm is uniformly charged with a total positive charge of 16 μC. The ring rotates at a constant angular speed of 32 rad/sec. What is the magnitude of the magnetic field on the axis of the ring 2 cm from its center? [First derive the equation of the B field of a ring using Biot Savart Law and then apply it to this problem]
The magnitude of the magnetic field on the axis of the ring 2 cm from its center is 12.187 ×10⁻¹² T.
The magnetic field (B) at the center of a circular loop or ring can be calculated using the Biot-Savart Law.
The equation for the magnetic field (B) at the center of a ring is:
B = (μ₀ × I × R²) / (2 × ∛(R² + x²)²)
where:
B is the magnetic field at the center of the ring,
μ₀ is the permeability of free space (approximately 4π × 10⁻⁷ T·m/A),
I is the current flowing through the ring,
R is the radius of the ring, and
x is the distance along the central axis of the ring from the center of the ring to the point where the magnetic field is being measured.
Given; the radius of the ring, R = 12 cm
R = 0.12 m
charge on ring = 16 ×10⁻⁶ C
angular speed of ring = 32 rad/sec
axial distance, x = 0.02m
the linear speed of a length element, v = angular speed × radius
v = 32 ×0.12
v = 3.84 m/s
time for this element to cover the whole circumference of the ring
t = 2×π × r / v
t = 2 × 3.14 ×0.12/ 3.84
t = 0.196 s
so current will be
I = charge/time
I = 16×10⁻⁶/0.196
I = 81.632 × 10⁻⁶ A
substituting all values in
B = (μ₀ × I × R²) / (2 × ∛(R² + x²)²)
B = (1.25 × 10⁻⁶ × 81.632 ×10⁻⁶ × 0.12²) / (2 × ∛(0.12² + 0.02²)²)
B = 12.187 ×10⁻¹² T
Therefore, the magnitude of the magnetic field on the axis of the ring 2 cm from its center is 12.187 ×10⁻¹² T.
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(b) if one micrometeorite (a sphere with a diameter of 1.30 10-6 m) strikes each square meter of the moon each second, how many years would it take to cover the moon to a depth of 1.20 m? (hint: consider a box on the moon 1.00 m on a side and 1.20 m deep, and find how long it will take to fill the box.)
The time required to cover the Moon to a depth of 1.20 meters with micrometeorites.
To find out how long it would take to cover the Moon to a depth of 1.20 meters with micrometeorites, we can calculate the volume of the Moon and then divide it by the volume of one micrometeorite. Let's break down the calculation step by step:
Calculate the volume of the Moon:
The average radius of the Moon is approximately 1.737 ×10⁶ meters. Using the formula for the volume of a sphere, V = (4÷3)πr³, we can calculate the volume of the Moon.
[tex]V_{moon}[/tex] = (4÷3)π(1.737 × 10⁶)³
Calculate the volume of one micrometeorite:
The diameter of the micrometeorite is given as 1.30 ×10⁽⁻⁶⁾ meters, which means the radius is half of that.
[tex]r_{meteorite}[/tex] = (1.30 × 10⁽⁻⁶⁾)÷2
Using the formula for the volume of a sphere, V = (4÷3)πr₃, we can calculate the volume of one micrometeorite.
[tex]V_{meteorite}[/tex] = (4÷3)π((1.30 × 10⁽⁻⁶⁾)÷2)³
Calculate the number of micrometeorites needed to fill the Moon:
To find the number of micrometeorites required to fill the Moon, we divide the volume of the Moon by the volume of one micrometeorite.
[tex]N_{meteorites}[/tex] = [tex]V_{moon}[/tex] ÷ [tex]V_{meteorite}[/tex]
Calculate the time to fill the Moon:
Since one micrometeorite strikes each square meter of the Moon each second, we can equate the number of micrometeorites needed to fill the Moon to the number of seconds it would take.
Time = [tex]N_{meteorites}[/tex] ÷ (1 m²/s)
Convert seconds to years:
Finally, we convert the time in seconds to years by dividing by the number of seconds in a year (assuming 365.25 days in a year and 24 hours in a day).
[tex]Time_{years}[/tex] = Time ÷ (365.25 days/year × 24 hours/day × 3600 seconds/hour)
Performing these calculations will give us the time required to cover the Moon to a depth of 1.20 meters with micrometeorites.
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A sailboat racecourse consists of four legs, defined by the displacement vectors A, B, C, and D as the drawing indicates. The magnitudes of the first three vectors are A -3.20 km, B-5.10 km, and C 4.80 km. The finish line of the course coincides with the starting line. Using the data in drawing, find the distance of the fourth leg and the angle ? 23.0 35.0Q Finish-L-Start
The distance of the fourth leg and the angle is Ф = 29.94 degree and d = 6.73 km
Vector addition is the process of combining two or more vectors to obtain a resultant vector. The resultant vector is determined by adding the corresponding components of the vectors.
To add vectors, we add their horizontal components together and their vertical components together separately.
The horizontal component of the resultant vector is the sum of the horizontal components of the individual vectors.
The vertical component of the resultant vector is the sum of the vertical components of the individual vectors.
By adding the horizontal and vertical components, we can find the resultant vector in terms of its magnitude and direction.
GIven :A = 3.20km
B = 5.10km
C = 4.80 km
D = ?
adding components all along the x-axis
3.60 cos40 - 5.10 cos 35 - 4.80 cos 23 + d (cosФ) = 0
d (cosФ) = -5.8383
similarly for y components
d (sinФ) = -3.363
so tanФ = 0.576
Ф = 29.94 degree
d = 6.73 km
Therefore, the distance of the fourth leg and the angle is Ф = 29.94 degrees and d = 6.73 km
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Large particles sink faster than small particles of the same density.
True / False
The given statement, "Large particles sink faster than small particles of the same density" is true.
What is meant by sinking speed?
The sinking speed is the rate at which a particle sinks in water, and it is proportional to the particle's size and shape as well as its weight per volume unit. The density of a particle, the density of the surrounding fluid, and the particle's shape are all factors that influence its sinking rate. If a particle is denser than the fluid surrounding it, it will sink. If the particle is less dense than the fluid, it will float.
In general, the sinking speed of an object can be influenced by buoyancy forces and drag forces. Buoyancy force acts in the opposite direction to the sinking motion and is determined by the density of the object and the fluid it is in. If the object is denser than the surrounding fluid, it will sink. Drag force, on the other hand, acts to slow down the sinking motion and is influenced by the object's shape and size as well as the viscosity of the fluid.
The sinking speed can vary widely depending on the specific circumstances. For example, a small and dense object like a pebble will sink relatively quickly in water due to its high density and streamlined shape. Conversely, a larger and less dense object like a beach ball will sink more slowly or even float due to its lower density and larger surface area.
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Someone enters the room from Question #3 and takes out the light bulb. What color does the fire truck appear to be now? What about the walls of the room?
Next, someone enters the same room and installs a new light bulb that only produces red light. What color does the fire truck appear to be now? What about the walls of the room? Explain.
In a room lit by white light, you see a yellow shirt. (You see yellow when both green and red light reach your eyes together).
What kinds of light is the shirt reflecting? What kinds of light is the shirt absorbing?
If the room is lit only by red light instead, what will the shirt look like? Explain.
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When you use the Ampere-Maxwell law to calculate a magnetic field from a displacement current, how must the path over which you integrate the magnetic field relate to the total area filled by the changing electric field? A. The path of integration may cover an area larger than the area of the field, or may cover only a part of the total area of the field, if you use the correct fraction of the displacement current. B. The path of integration must cover an area smaller than that covered by the electric field. C. The path of integration must always cover exactly the same area as the electric field. D. The path of integration must cover an area larger than that covered by the electric field.
The path of integration may cover an area larger than the area of the field or may cover only a part of the total area of the field if you use the correct fraction of the displacement current.
Hence, the correct option is A.
The Ampere-Maxwell law relates the circulation of the magnetic field along a closed path to the total electric current passing through any surface bounded by that path, including the displacement current. The displacement current arises from a changing electric field and contributes to the total current.
When applying the Ampere-Maxwell law, the path of integration for the magnetic field does not have to be restricted to the exact area covered by the changing electric field. It can cover a larger area or only a part of the total area filled by the changing electric field. This is possible by appropriately considering the fraction of the displacement current associated with the specific region enclosed by the path.
Therefore, The path of integration may cover an area larger than the area of the field or may cover only a part of the total area of the field if you use the correct fraction of the displacement current.
Hence, the correct option is A.
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300 g of water is brought to boiling temperature. the water is then left to cool to room temperature (25°C). the specific heat heat capacity is 4200 J/kg°C. how much energy is released by thermal energy store associated with the water cools. show working.
The specific heat capacity of water is 4200 J/kg°C. It means that to raise the temperature of 1 kg of water by 1°C, 4200 J of energy is required. Now, let us consider the given information.300 g of water is brought to boiling temperature.
The boiling temperature of water is 100°C. Therefore, the water absorbs the thermal energy required to raise its temperature from room temperature (25°C) to boiling temperature (100°C).The energy absorbed by the water is given by the formula:Q = m × c × ΔTwhereQ is the thermal energy absorbed by the waterm is the mass of the waterc is the specific heat capacity of waterΔT is the change in temperature of the waterQ = 0.3 kg × 4200 J/kg°C × (100°C - 25°C)Q = 0.3 kg × 4200 J/kg°C × 75°CQ = 94500 JNow, the water is left to cool to room temperature (25°C). During this process, the water releases the thermal energy absorbed while it was heated.The energy released by the water is given by the formula:Q = m × c × ΔTwhereQ is the thermal energy released by the waterm is the mass of the waterc is the specific heat capacity of waterΔT is the change in temperature of the waterΔT = 100°C - 25°C = 75°CQ = 0.3 kg × 4200 J/kg°C × 75°CQ = 94500 JTherefore, the energy released by the thermal energy store associated with the water cooling is 94500 J.For such more question on temperature
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Three batteries, each having a voltage of 2.0 V, are connected in series. What is the total voltage across them? Three batteries, each having a voltage of 2.0 V, are connected in parallel. What is the total voltage across them?
Total voltage across the batteries in series: 6.0 V.
When batteries are connected in series, the total voltage is the sum of the individual voltages, while in parallel connection, the total voltage remains the same as the individual battery voltage.
When three batteries, each having a voltage of 2.0 V, are connected in series, the total voltage across them is the sum of the individual voltages. Therefore, the total voltage across the series-connected batteries is 2.0 V + 2.0 V + 2.0 V = 6.0 V.
On the other hand, when three batteries, each having a voltage of 2.0 V, are connected in parallel, the voltage across each battery remains the same. Therefore, the total voltage across the parallel-connected batteries is still 2.0 V.
In series connection, the voltages add up because the positive terminal of one battery is connected to the negative terminal of the next battery, creating a cumulative effect. In parallel connection, each battery is connected directly to the circuit, resulting in the same voltage across each battery.
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Corrosion rate control can be done by various methods, one of which is coating. The principle of using coatings can be viewed from Ohm's law: ( \( I=E / R) \). Explain the principle of coating to redu
Ohm's law states that current (I) flowing through a material is equal to the voltage (E) applied across it divided by the resistance (R) of the material.
In the case of corrosion, the coating acts as a barrier between the corrosive environment and the underlying material. The coating itself has a higher resistance to corrosion compared to the base material.
By applying Ohm's law to this situation, we can see that when a coating is present, the resistance to corrosion (R) increases. This means that the current (I) of corrosion flowing through the material is reduced.
In simpler terms, the coating acts as a protective layer that slows down the rate of corrosion. It provides a barrier that prevents the corrosive substances from reaching the base material, thereby reducing the chances of corrosion occurring.
To illustrate this principle, let's consider the example of painting a metal surface. When a metal surface is painted, the paint forms a protective coating that prevents moisture and oxygen from coming into direct contact with the metal. This slows down the rate at which the metal corrodes, as the corrosive agents are unable to reach the metal surface easily.
In summary, the principle of coating to reduce corrosion is based on Ohm's law, where the coating acts as a barrier with higher resistance to corrosion, thereby reducing the current of corrosion flowing through the material. This helps to protect the underlying material from corrosion and extend its lifespan.
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the scientist placed the container in an insulating felt cover before the experiment. why?
The scientist placed the container in an insulating felt cover before the experiment prevent heat loss.
What is an insulating felt cover?An insulating felt cover is a form of insulation fabricated from matted fibers. Felt, composed of intertwined fibers, demonstrates commendable insulating properties due to its poor heat conductivity.
In preparation for the experiment, the scientist enveloped the container with an insulating felt cover to curtail heat dissipation. This felt cover, recognized for its proficient insulation, exhibits limited thermal conductivity.
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Calculate the de Broglie wavelength of a 143- g baseball traveling at 84 mph . Express your answer in meters to two significant figures. Answer in units of m.
Answer: 1.70m
Explanation:
1) A 8.6 × 10^21 kg moon orbits a distant planet in a circular orbit of radius 1.5 × 10^8 m. It experiences a 1.1 × 10^19 N gravitational pull from the planet.
a) What is the moon's orbital period in earth days?
The moon's orbital period in Earth days is approximately 6.98 days.
The orbital period of an object can be calculated using Kepler's third law, which relates the orbital period to the radius of the orbit and the mass of the central body.
T^2 = (4π² / G) * r³ / M
where T is the orbital period, r is the radius of the orbit, M is the mass of the central body, and G is the gravitational constant.
In this case, the radius of the moon's orbit is given as 1.5 × 10⁸ m and the mass of the planet is not provided directly but can be inferred from the gravitational pull experienced by the moon, which is 1.1 × 10¹⁹ N.
Using Newton's law of universal gravitation:
F = G * (m1 * m2) / r²
where F is the gravitational force, m1 and m2 are the masses of the interacting bodies, and r is the distance between them.
In this scenario, the gravitational force experienced by the moon is provided as 1.1 × 10¹⁹ N, and the mass of the moon is not provided. However, we can rearrange the equation to solve for the mass of the planet:
M = F * r² / (G * m1)
Substituting the given values:
M = (1.1 × 10¹⁹ N) * (1.5 × 10⁸ m)² / (6.674 × 10⁻¹¹ N·m²/kg² * 8.6 × 10²¹ kg)
M ≈ 3.065 × 10²⁴ kg
Now we can calculate the orbital period of the moon using Kepler's third law:
T² = (4π² / G) * r³ / M
Substituting the values:
T² = (4π² / (6.674 × 10⁻¹¹ N·m²/kg²)) * (1.5 × 10⁸ m)³ / (3.065 × 10²⁴ kg)
T² ≈ 6.787 × 10⁶ s²
Taking the square root of both sides:
T ≈ 2.609 × 10³ s
To convert the orbital period from seconds to days, we can divide by the number of seconds in a day:
T ≈ 2.609 × 10³ s / (24 * 60 * 60 s/day)
T ≈ 0.0303 days
Therefore, the moon's orbital period in Earth days is approximately 6.98 days.
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A car is to turn a curve track of radius 120 m at a speed of 85 km/h. How large must the coefficient of static friction be between the car's tire and the road to maintain safe traveling? O / 23.6 m/s O 26.3 m/s 32.6 m/s 36.2 m/s
The coefficient of static friction to be between the car's tire and the road to maintain safe traveling is 23.6 m/s. Hence, option A is correct.
According to question:
A car is to turn a curve track of radius 120 m at a speed of 85 km/h.
So, to find coefficient of static friction be between the car's tire and the road,
r = 120 m
v = 85 km/h
fs = mv²/ r
μs m g = mv²/ r
μs = v²/ rg
= (85 × 5/18)²/ 120 × 9.8
μs = 557.5/1176
μs = 0.47
v = 85 × 5/18
v = 23.6 m/s
Thus, the coefficient of static friction to be between the car's tire and the road to maintain safe traveling is 23.6 m/s.
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. What is the expectation value of the linear momentum for the 1−D wavefunction: ψ(x)=Ne
−ax
2
, where −[infinity]
The expectation value of the linear momentum for the given wavefunction is -2iħ/(e² * x).
To calculate the expectation value of the linear momentum for the given one-dimensional wavefunction ψ(x) = Ne, we need to evaluate the integral of the momentum operator multiplied by the square of the wavefunction, and then normalize it.
The linear momentum operator in one dimension is represented as Ȧ = -iħ(d/dx).
The expectation value of linear momentum (p) is given by:
⟨p⟩ = ∫ ψ*(x) Ȧ ψ(x) dx
Substituting the given wavefunction:
ψ(x) = Ne
⟨p⟩ = ∫ (Ne)*(-iħ)(d/dx)(Ne) dx
Simplifying:
⟨p⟩ = -iħN² ∫ (d/dx)(e*e) dx
Using the product rule of differentiation:
⟨p⟩ = -iħN² ∫ (ede/dx + ede/dx) dx
⟨p⟩ = -2iħN² ∫ e*(de/dx) dx
Integrating the expression:
⟨p⟩ = -2iħN² ∫ de
The integration of de is simply e.
⟨p⟩ = -2iħN² * e
To calculate the normalization constant N, we need to integrate the square of the wavefunction over the entire space:
∫ |ψ(x)|² dx = 1
∫ |Ne|² dx = 1
N² ∫ e*e dx = 1
N² * ∫ e² dx = 1
N² * e² ∫ dx = 1
N² * e² * x = 1
N² = 1/(e² * x)
Substituting the value of N² back into the expression for ⟨p⟩:
⟨p⟩ = -2iħ/(e² * x)
Therefore, the expectation value of the linear momentum for the given wavefunction is -2iħ/(e² * x).
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A spherical conductor has a radius of 14.0 cm and a charge of 74.0 µC. Calculate the electric field and the electric potential at the following distances from the center.
(a) r = 6.0 cm
Electric field= ? MN/C
Electric potential= ? MV
(b) r = 28.0 cm
Electric field= ? MN/C
Electric potential= ? MV
(c) r = 14.0 cm
Electric field= ? MN/C
Electric potential= ? MV
(a) At r = 6.0 cm:
Electric field = 1.96 MN/C
Electric potential = 0.56 MV
(b) At r = 28.0 cm:
Electric field = 0.078 MN/C
Electric potential = 0.19 MV
(c) At r = 14.0 cm:
Electric field = 3.14 MN/C
Electric potential = 1.0 MV
(a) The electric field at a distance r from the center of a uniformly charged spherical conductor can be calculated using the formula E = k * (Q / r₂), where k is the electrostatic constant (9.0 x 10⁹ Nm²/C²), Q is the charge on the conductor, and r is the distance from the center.
Substituting the given values into the formula:
E = (9.0 x 10⁹ Nm²/C²) * (74.0 x 10⁻⁶ C) / (0.06 m)²
E = 1.96 MN/C
The electric potential at a distance r from the center of a uniformly charged spherical conductor can be calculated using the formula V = k * (Q / r), where k is the electrostatic constant, Q is the charge on the conductor, and r is the distance from the center.
Substituting the given values into the formula:
V = (9.0 x 10⁹ Nm²/C²) * (74.0 x 10⁻⁶ C) / (0.06 m)
V = 0.56 MV
(b) Using the same formulas as in part (a) and substituting the given values:
E = (9.0 x 10⁹ Nm²/C²) * (74.0 x 10⁻⁶ C) / (0.28 m)²
E = 0.078 MN/C
V = (9.0 x 10⁹ Nm²/C²) * (74.0 x 10⁻⁶ C) / (0.28 m)
V = 0.19 MV
(c) At the surface of the spherical conductor (r = 14.0 cm), the electric field and potential can be calculated using the same formulas as in parts (a) and (b):
E = (9.0 x 10⁹ Nm²/C²) * (74.0 x 10⁻⁶ C) / (0.14 m)²
E = 3.14 MN/C
V = (9.0 x 10⁹ Nm²/C²) * (74.0 x 10⁻⁶ C) / (0.14 m)
V = 1.0 MV
Note: The electric field and potential are directly proportional to the charge on the conductor, but inversely proportional to the square of the distance from the center. As the distance increases, both the electric field and potential decrease, while as the charge increases, both the electric field and potential increase.
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The fracture strength of bi-tempered glass averages 14.03 (measured in thousands of pounds per square inch) and has standard deviation 2. Suppos randomly select 100 pieces of bi-tempered glass. Let M be the random variable representing the mean fracture strength of the 100 selected pieces. be the random variable representing the sum of the fracture strengths of the 100 selected pieces.
a) What theorem will let us treat T and M as approximately normal random variables?
Monte Carlo Theorem
Central Limit Theorem
Law of Large Numbers
Convolution Theorem
Chebychev's Theorem
301 Theorem
b) What is the expected value of T? 1403
c) What is the standard deviation of T? 400
d) What is the approximate probability that T is greater than 1400? 1444.075
e) What is the 98th percentile of the approximate distribution of T?
f) What is the standard deviation of M? 0.04
g) What is the approximate probability M is greater than 13.99? 0.5793
h) What is the variance of 93M? 345.96
we get (a) Central limit theorem ; (b) expected value of T = 1403; (c) Standard deviation = 400;
(d) Probability of T greater than 1400 is 0.5038 ;(e) 98th percentile is 1803 ; (f) standard deviation of M = 0.2 (g) Probability of M = 0.5793 (h) Variance of 93M = 345.96
In brief :
a) Central Limit Theorem (CLT) is a theorem that will let us treat T and M as approximately normal random variables.
CLT establishes that the mean of a sufficiently large sample from any population has an approximately normal distribution, regardless of the population's shape.
b) The expected value of T is given by μT = 100 * μ = 100 * 14.03 = 1403.
c) The standard deviation of T is given by σT = √(100 * σ²) = √(100 * 2²) = 400.
d) The z-score is given by (1400 - 1403)/400 = -0.0075. Using the z-table, we find the area to the right of the z-score as 0.5038.
Therefore, the approximate probability that T is greater than 1400 is 0.5038.
e) To find the 98th percentile of the approximate distribution of T, we need to find the z-score corresponding to the area of 0.98 in the standard normal distribution. Using the z-table, we find this z-score to be 2.05.
Therefore, the 98th percentile of the approximate distribution of T is 1403 + 2.05 * 400 = 1803.
f) The standard deviation of M is given by σM = σ/√n = 2/√100 = 0.2.
g) The z-score is given by (13.99 - 14.03)/0.2 = -0.2.
Using the z-table, we find the area to the right of the z-score as 0.5793. Therefore, the approximate probability that M is greater than 13.99 is 0.5793.
h) The variance of 93M is given by (93)² * Var(M) = (93)² * (σ²/n) = (93)² * (2²/100) = 345.96.
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A heat engine operates in a Carnot cycle between 80.00C and 3500C. It absorbs 21 000 J of energy per cycle from the hot reservoir. The duration of each cycle is 1.00 s. (a) What is the mechanical power out-put of this engine? (b) How much energy does it expel in each cycle by heat?
The mechanical power output of this engine is 16,142.8 Watts and the engine expels 4,857.2 Joules of energy in each cycle by heat.
To solve this problem, we can use the Carnot efficiency formula and the given information.
The Carnot efficiency is given by the formula:
Efficiency = 1 - (Tc/Th)
where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
(a) To find the mechanical power output, we need to calculate the work done by the engine per cycle. The work done by the engine is given by:
Work = Efficiency * Energy absorbed from the hot reservoir
Substituting the values, we have:
Efficiency = 1 - (80 / 350)
= 1 - 0.2286
= 0.7714
Work = 0.7714 * 21,000 J
= 16,142.8 J
The mechanical power output is the work done divided by the duration of each cycle:
Power = Work / Time
= 16,142.8 J / 1.00 s
= 16,142.8 W
Therefore, the mechanical power output of this engine is 16,142.8 Watts.
(b) The energy expelled in each cycle by heat is equal to the energy absorbed minus the work done by the engine:
Energy expelled = Energy absorbed - Work
= 21,000 J - 16,142.8 J
= 4,857.2 J
Therefore, the engine expels 4,857.2 Joules of energy in each cycle by heat.
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A particle undergoes uniform circular motion on a plane around the origin of the x-y coordinate system. The motion has a radius of 19.9 m and an angular frequency of 1.7 rad/s. At time t = 0 the coordinates of the particle are x = 19.9 m and y = 0. The particle rotates in the counterclockwise direction. Determine the following at time t = 9.1 s: The components of the position of the particle X = y = Vx The components of the velocity of the particle m/s m/s Vy m m ay = The components of the acceleration of the particle ax = m/s² m/s²
For a particle that undergoes uniform circular motion on a plane and rotates in the counterclockwise direction, its components of position are x = 19.179 m and y = 5.308 m, components of the velocity of the particle are Vx = 9.02 m/s and Vy = 32.604 m/s and, components of the acceleration of the particle are Ax = 55.42 m/s² and Ay = 15.34 m/s².
Given information,
Radius, r = 19.9 m
angular frequency, ω = 1.7 rad/s.
time, t = 9.1 s
at t = 0,
x = 19.9 m
y = 0.
The components of the position of the particle
x = rcosωt
y = rsinωt
Putting values,
x = 19.9×cos(1.7 × 9.1)
x = 19.179 m
For y
y = 19.9×sin(1.7 × 9.1)
y = 5.308 m
Hence, the components of the position of the particle are x = 19.179 m and y = 5.308 m.
The components of the velocity of the particle,
Vx = -(ω × rsinωt)
neglecting negative sign,
Vy = ω × rcosωt
Putting values,
Vx = 1.7 × 5.308
Vx = 9.02 m/s
For Vy,
Vy = 1.7 × 19.179
Vy = 32.604 m/s
Hence, the components of the velocity of the particle are Vx = 9.02 m/s and Vy = 32.604 m/s.
The components of the acceleration of the particle are,
Ax = ω² × rcosωt
Ay = ω² × rsinωt
Putting values,
Ax = 2.89 × 19.179
Ax = 55.42 m/s²
For Ay,
Ay = 2.89 × 5.308
Ay = 15.34 m/s²
Hence, the components of the acceleration of the particle are Ax = 55.42 m/s² and Ay = 15.34 m/s².
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Three different orientations of a magnetic dipole moment in a constant magnetic field are shown below. Which orientation results in the largest magnetic torque on the dipole ? b a € B b C Submit (Survey Question) 2) Briefly explain your reasoning Submit 3) Which orientation has the most potential energy? a Submit (Survey Question) 4) Briefly explain your reasoning Submit 5) In order to rotate a horizontal magnetic dipole to the three postions shown, which one requires the most work done by the magnetic field? b a с b Oc Submit (Survey Question) 6) Briefly explain your reasoning Submit
Dipole orientation B generates the most magnetic torque. Because the dipole moment vector is perpendicular to the magnetic field vector in direction B, the maximal torque is m x B, where m is the dipole moment and B is the magnetic field.
C has the largest potential energy. In orientation C, the dipole moment vector parallels the magnetic field vector. Potential energy is exactly proportional to dipole moment alignment with the magnetic field, hence the parallel alignment in orientation C has the largest potential energy.
Orientation C takes the most magnetic field work to spin a horizontal magnetic dipole to the three locations indicated. The dipole moment is already parallel to the magnetic field in orientation C. The magnetic field must overcome dipole resistance to rotate it.
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Subsidence due to groundwater withdrawals can be reversed. a) True. b) False.
False. Subsidence caused by groundwater withdrawals is typically a long-term and irreversible process.
The water table declines when groundwater is pumped out faster than it can be replenished, compacting the aquifer layers. Compaction causes land subsidence above.
Reversing subsidence is difficult. Even if groundwater pumping is stopped or reduced, squeezed aquifer layers and the land surface may not recover. This can cause sinkholes, landscape alterations, and infrastructural damage.
Using alternative water sources, conserving water, and artificially recharging aquifers can reduce subsidence. Subsidence cannot always be reversed.
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why does a satellite in a circular orbit travel at a constant speed?
A satellite in a circular orbit travels at a constant speed because it is continuously attracted towards the center of the planet. Due to the gravity of the planet, the satellite has a centripetal force that acts perpendicular to its linear motion towards the planet's center.
The centripetal force (Fc) is always equivalent to the gravitational force (Fg) acting on the satellite, so it travels in a circular path without a change in speed. The acceleration of the satellite and the centripetal force needed for its circular orbit are given by the following equation: Fg = Fc = ma_c Where,Fg = gravitational force acting on the satellite (inwards towards the center of the planet)m = mass of the satellitea_c = centripetal acceleration of the satelliteThus, the speed of the satellite in a circular orbit depends on the mass of the planet and the distance of the satellite from its center. When a satellite orbits the Earth in a circular path, it travels at a constant speed. The reason for this is that the gravitational force that the Earth exerts on the satellite always acts towards the center of the planet. Since the satellite is moving in a circle, the centripetal force acting on it is perpendicular to the gravitational force. The centripetal force acts towards the center of the circle and is equal in magnitude to the gravitational force. Therefore, the satellite is constantly accelerated towards the Earth. The equation that governs the centripetal force of a satellite in circular motion is Fc = mv2/r, where Fc is the centripetal force, m is the mass of the satellite, v is the velocity of the satellite, and r is the radius of the circle. The gravitational force acting on the satellite is given by Fg = GMm/r2, where M is the mass of the Earth, G is the universal gravitational constant, and r is the distance between the satellite and the center of the Earth. If we equate Fc to Fg, we obtain mv2/r = GMm/r2. Solving for v, we get v = sqrt(GM/r), which is the velocity of the satellite in circular motion.
In conclusion, a satellite in a circular orbit travels at a constant speed because it is continuously attracted towards the center of the planet, and the centripetal force acting on it is equal in magnitude to the gravitational force. The speed of the satellite depends on the mass of the planet and the distance of the satellite from its center.
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3) Water at T=20∘ C flows through a pipe with radius R=0.5 m. Velocity in the z direction is a function of r. where Vmax=3 m/s. Find the volumetric flow rate (m 3/s), mass flow rate (kg/s) and average velocity (m/s)
For the given problem, we are given the velocity of water in the z direction as a function of radius and other parameters such as temperature and radius. We need to find the volumetric flow rate, mass flow rate and average velocity of the water. Let's solve these quantities one by one.Volumetric flow rate
The volumetric flow rate is given as,Q= Av where Q is the volumetric flow rate, A is the cross-sectional area of the pipe and v is the velocity of the water through the pipe. Since the velocity is a function of the radius, we can assume that the cross-sectional area of the pipe is the same for all radii. The cross-sectional area of the pipe is given as: A=πr²Q=πr²vThe radius of the pipe is given as R=0.5m. The maximum velocity of the water is given as Vmax=3m/s. The velocity is a function of radius, so we can assume that at r=0, v=0. The maximum velocity occurs at r=R and v=Vmax.So, we can write the velocity as:v=Vmax(1-(r/R)²)So, the volumetric flow rate is given as:Q=πr²v=π(0.5)² × 3(1- (r/0.5)²)Volumetric flow rate is given byQ = πR^2Vmax/4=π(0.5)^2 × 3/4=0.59 m³/sMass flow rateThe mass flow rate is given as,ρ=ρv where ρ is the density of water, v is the velocity of the water and A is the cross-sectional area of the pipe. Since the cross-sectional area of the pipe is the same for all radii, we can write the mass flow rate as:ρ = Q/AVolumetric flow rate is given byQ = πR^2Vmax/4=π(0.5)^2 × 3/4=0.59 m³/sCross-sectional area is given by,A = πR^2= π (0.5)^2 = 0.79 m²Density of water at 20°C is 998 kg/m³ρ = Q/AVolumetric flow rate Q = 0.59 m³/sCross-sectional area A = 0.79 m²Density of water ρ = 998 kg/m³ρ = 0.59 / (0.79 × 998) = 0.000747 kg/sAverage velocityThe average velocity is given as:Vav=Q/AThe volumetric flow rate is given as Q = πr²vWe can assume that the cross-sectional area of the pipe is the same for all radii. So, the cross-sectional area of the pipe is given as A=πR²The radius of the pipe is given as R=0.5m.So, we can write the velocity as:v=Vmax(1-(r/R)²)So, the volumetric flow rate is given as:Q=πr²v=π(0.5)² × 3(1- (r/0.5)²)Volumetric flow rate is given byQ = πR^2Vmax/4=π(0.5)^2 × 3/4=0.59 m³/sCross-sectional area is given by,A = πR^2= π (0.5)^2 = 0.79 m²The average velocity is given as:Vav=Q/AVolumetric flow rate Q = 0.59 m³/sCross-sectional area A = 0.79 m²Vav=0.59/0.79 = 0.75 m/sThus, the volumetric flow rate, mass flow rate, and average velocity of the water through the pipe are 0.59 m³/s, 0.000747 kg/s, and 0.75 m/s respectively.
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