If the pressure in the pipe is 8940 n/m° and in the constricted section is 6705 n/m, the rate at which oil is flowing through the pipe is 2.74 m^3/s.
The volume flow rate (Q) can be expressed as:
Q = Av
We can use the continuity equation to relate the velocities at the two different cross-sectional areas:
A1v1 = A2v2
Where:A1 and A2 are the cross-sectional areas at the two different points in the pipe
v1 and v2 are the fluid velocities at the two different points in the pipe
We can rearrange this equation to solve for v1 in terms of v2:
v1 = (A2/A1) * v2
Now we can use Bernoulli's equation to relate the pressure and velocity at the two different points in the pipe:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
We can rearrange this equation to solve for v2:
v2 = sqrt(2(P1 - P2) / ρ + v1^2)
Now we can plug in the given values to find v2:
v1 = ?
A1 = (π/4)(0.945 m)^2 = 0.664 m^2
A2 = (π/4)(0.567 m)^2 = 0.100 m^2
ρ = 821 kg/m^3
P1 = 8940 N/m^2
P2 = 6705 N/m^2
v2 = sqrt(2(8940 - 6705) / 821 + ((0.100/0.664) * v1)^2)
We can simplify this equation by assuming that the velocity at the wider section of the pipe (v1) is much smaller than the velocity at the constriction (v2), due to the principle of continuity. Therefore, we can neglect the second term in the equation, giving:
v2 = sqrt(2(8940 - 6705) / 821) = 11.98 m/s
Now we can use the continuity equation to find the volume flow rate:
Q = A1v1 = A2v2 = (π/4)(0.945 m)^2 * 11.98 m/s =2.74 m^3/s
Therefore, the rate at which oil is flowing through the pipe is 2.74 m^3/s.
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An object has an access of 50 electrons what is the charge on the object
The charge on the object is 8.01 x 10^-18 Coulombs.
Q = Ne
where Q is the charge, N is the number of excess or deficit electrons on the object, and e is the elementary charge, which is approximately 1.602 x 10^-19 Coulombs.
Q = 50 x 1.602 x 10^-19 C/electron
Q = 8.01 x 10^-18 C
A charge is a fundamental property of matter that describes the amount of electric energy that a particle possesses. It is a scalar quantity, meaning it has only magnitude and no direction. The charge can be either positive or negative and is measured in Coulombs (C).
Charged particles can interact with each other through electromagnetic force, which is one of the four fundamental forces of nature. Like charges repel each other while opposite charges attract. Electric fields are generated by charged particles, and the force they exert on other charged particles is proportional to the magnitude of their charges.
A charge is conserved in isolated systems, meaning that the total amount of charge in a closed system cannot be created or destroyed, but can only be transferred from one object to another.
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10. You weigh yourself on a scale and find that your weight is about 126 pounds.
a) Multiply your weight in pounds times 4.45. This is your weight in Newtons.
b) What is your mass in kilograms?
c) If you went to the moon, how much would you weigh?
(a) Your weight in Newtons is 560.7 N.
(b) Your mass in kilograms is 57.12 kg.
(c) Your weight on the moon is 93.45 N.
What is the measure of your weight in Newtons?a) To convert your weight from pounds to Newtons, we can use the conversion factor that 1 pound is equal to 4.45 Newtons.
Therefore, your weight in Newtons would be:
126 pounds x 4.45 Newtons/pound = 560.7 Newtons
b) To find your mass in kilograms, we can use the formula:
mass (kg) = weight (N) / acceleration due to gravity (m/s²)
The acceleration due to gravity on Earth is approximately 9.81 m/s². Therefore, your mass in kilograms would be:
mass (kg) = 560.7 N / 9.81 m/s² = 57.12 kg
c) The gravitational force between two objects depends on their masses and the distance between them. The gravitational force on the moon is about 1/6th the gravitational force on Earth.
Therefore, if you went to the moon, your weight would be about 1/6th of what it is on Earth.
Your weight on the moon would be:
126 pounds / 6 = 21 pounds
To convert this to Newtons, we can use the conversion factor:
1 pound = 4.45 Newtons
So your weight on the moon in Newtons would be:
21 pounds x 4.45 Newtons/pound = 93.45 Newtons
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how much voltage is available to operate an overvoltage relay 59g when connected across the grounding resistor? what is the multiple of pickup if 59g minimum operating value is 5.4 v.
The voltage available to operate an overvoltage relay 59G when connected across the grounding resistor is 54 V. The multiple of pickup if 59G minimum operating value is 5.4 V is 10.
An overvoltage relay is a type of protective relay that is used to detect overvoltage conditions on an electrical network. The purpose of this relay is to detect if the voltage on the network exceeds a certain level, which could result in damage to equipment or electrical systems connected to the network. The overvoltage relay is connected to the electrical network in such a way that it monitors the voltage level on the network and is capable of tripping a circuit breaker or other protective device if the voltage level exceeds a preset value.
The voltage available to operate an overvoltage relay 59G when connected across the grounding resistor is 54 V. This is because the grounding resistor limits the voltage to a maximum of 54 volts. If the voltage level on the network.
The multiple of pickup if 59G minimum operating value is 5.4 V is 10. This means that the overvoltage relay will trip if the voltage on the network exceeds 54 volts. If the voltage level on the network is between 5.4 volts and 54 volts, the overvoltage relay will not trip, since the voltage is below the pickup level. However, if the voltage level on the network exceeds 54 volts, the overvoltage relay will trip and disconnect the circuit.
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List down the forces you exert from the moment you wake up in the morning to the time you go to sleep. At least five and write or draw in a short bond paper
Here are at least five forces that people may exert from the moment they wake up in the morning to the time they go to sleep:
Gravitational force: This force is exerted on our bodies from the moment we wake up in the morning, keeping us grounded to the surface of the earth. Muscular force: As we move and perform daily activities, our muscles exert force to lift, carry, push, or pull objects. Frictional force: This force is exerted when we walk or run, helping us grip the ground and move forward. Air resistance force: As we move through the air, our bodies and clothing experience air resistance, which can affect our movement and speed. Electrical force: Our nervous system relies on electrical signals to communicate with our muscles and organs, which help us perform various actions throughout the day. These forces may vary depending on the activities a person engages in throughout their day, but they all play a role in the functioning of our bodies and the world around us.
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A runner with a mass of 90 kg accelerates from 0 to 8 m/s in 4 s. Find the net force on the runner using the alternate form of newton's second law
Answer: 180 Newtons.
Explanation:
The alternate form of Newton's second law states that force (F) is equal to mass (m) multiplied by acceleration (a):
F = m x a
In this problem, the mass of the runner (m) is 90 kg, and the acceleration (a) can be found by dividing the change in velocity (8 m/s) by the time interval (4 s):
a = (8 m/s) / (4 s) = 2 m/s^2
Substituting the values into the equation for force:
F = 90 kg x 2 m/s^2
F = 180 N
Therefore, the net force on the runner is 180 Newtons.
3. Why don't solar eclipses happen every
month? sc.8.E.5.9
A
B
C
D
Earth's orbit around the Sun is at an
angle, which keeps Earth out of the
Moon's shadow.
Earth's tilt on its rotation axis keeps it
from falling into the Moon's shadow
during most months.
The Moon's orbit is irregular, and most
of the time the Moon is too far away to
cast a shadow on Earth.
The Moon's orbit is tilted compared
to Earth's orbit, so Earth is not in the
Moon's shadow most months.
Answer:
D: The Moon's orbit is tilted compared
to Earth's orbit, so Earth is not in the
Moon's shadow most months.
A disk with radius R has uniform surface charge density σ.
Part A
By regarding the disk as a series of thin concentric rings, calculate the electric potential V at a point on the disk's axis a distance x from the center of the disk. Assume that the potential is zero at infinity. (Hint: Use the result that potential at a point on the ring axis at a distance x from the center of the ring is V=14πϵ0Qx2+a2√ where Q is the charge of the ring. )
Express your answer in terms of the given quantities and appropriate constants.
Part B
Calculate −∂V/∂x.
Express your answer in terms of the given quantities and appropriate constants
Part A: The electric potential V at a point on the disk's axis a distance x from the center of the disk is given by:
V = σ/2ε₀ × [tex](R^{2}/(x^{2} +R^{2} )^{1/2})[/tex]
Part B: After calculating for −∂V/∂x we get:
-∂V/∂x = σR²x/2ε₀[tex](x^{2}+R^{2})^{3/2}[/tex]
Part A:
The disc can be split into a number of thin, concentric rings in order to compute the electric potential V at a point on its axis that is located x distance from the disk's centre.
Each ring's potential is determined by:
[tex]V_{ring}[/tex] = 1/4πε₀ × ([tex]Q_{ring}[/tex] / [tex](x^{2} +R^{2} )^{1/2}[/tex])
where
[tex]Q_{ring}[/tex] is the charge of the ring and
ε₀ is the permittivity of free space.
Since
the disk has uniform surface charge density σ, the charge on each ring is given by:
[tex]Q_{ring}[/tex] = σ × 2πr × dr
where
r is the radius of the ring and
dr is its thickness.
By substituting [tex]Q_{ring}[/tex] into the expression for [tex]V_{ring}[/tex], we get:
[tex]V_{ring}[/tex] = 1/4πε₀ × (σ × 2πr × dr / [tex](x^{2} +R^{2} )^{1/2}[/tex])
By integrating across all the rings, it is possible to get the total potential V at any point along the axis of the disc:
V = ∫V_ring
V = ∫(1/4πε₀ × (σ x 2πr × dr / [tex](x^{2} +R^{2} )^{1/2}[/tex])
V = σ/2ε₀ × ∫(r / [tex](x^{2} +R^{2} )^{1/2}[/tex]) dr from 0 to R
By evaluating the integral and simplifying, we get:
V = σ/2ε₀ × [[tex](R^{2}/(x^{2} +R^{2} )^{1/2})[/tex] - [tex](0/(x^2+0^2)^{1/2})[/tex]]
V = σ/2ε₀ × [tex](R^{2}/(x^{2} +R^{2} )^{1/2})[/tex]
Therefore, the electric potential V at a point on the disk's axis a distance x from the center of the disk is given by:
V = σ/2ε₀ × [tex](R^{2}/(x^{2} +R^{2} )^{1/2})[/tex]
Part B:
To find the value of −∂V/∂x,
The derivative of the equation for V with regard to x must be taken:
∂V/∂x = -σR²x/2ε₀[tex](x^{2}+R^{2})^{3/2}[/tex]
Hence, the expression for −∂V/∂x is:
-∂V/∂x = σR²x/2ε₀[tex](x^{2}+R^{2})^{3/2}[/tex]
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Which statement describes the location of an earthquake’s epicenter?
It is located using a single set of data.
It is determined by the Mercalli Scale.
It is determined by the arrival time of surface waves.
It is located at the point directly above the focus
Answer:
it is located at the point directly above the focus
Answer:
D
Explanation:
#10. When the weightlifter in question #9 stands, his feet make contact with the floor over an area of 250 square centimeters. What is the area of contact between his feet and the floor in *square meters (picture of question 9 for reference)
The weight is 1372 N
How do we calculate the weight of an object on earth?The weight of an object on Earth can be calculated by using the following formula:
Weight = Mass x Acceleration due to Gravity
Where:
Weight is measured in Newtons (N)
Mass is measured in kilograms (kg)
Acceleration due to Gravity is a constant value
Weight = mass * acceleration due to gravity
Weight = 140 Kg * 9.8
= 1372 N
To calculate the weight of an object on Earth, you need to know its mass.
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you toss a racquetball directly upward and then catch it at the same height you released it 1.56 s later. assume air resistance is negligible. (a) what is the acceleration of the ball while it is moving upward? magnitude m/s2 direction ---select--- (b) what is the acceleration of the ball while it is moving downward? magnitude m/s2 direction ---select--- (c) what is the acceleration of the ball while it is at its maximum height? magnitude m/s2 direction ---select--- (d) what is the velocity of the ball when it reaches its maximum height? magnitude m/s direction ---select--- (e) what is the initial velocity of the ball? magnitude m/s direction ---select--- (f) what is the maximum height that the ball reaches?
The correct answer for all questions are: a), b) & c) will be acceleration due to gravity g=9.8 m/s^2, d) 0 m/s, e) 7.644 m/s, f) 2.977 m
A) The acceleration of the ball while it is moving upward is the acceleration due to gravity, which is 9.8 m/s2 in the downward direction.
B) The acceleration of the ball while it is moving downward is also the acceleration due to gravity, which is 9.8 m/s2 in the downward direction.
C) The acceleration of the ball while it is at its maximum height is still the acceleration due to gravity, which is 9.8 m/s2 in the downward direction.
D) The velocity of the ball when it reaches its maximum height is 0 m/s, because it has stopped moving upward and is about to start moving downward.
E) The initial velocity of the ball can be found using the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.
Since the ball is caught at the same height it was released, the final velocity is equal to the initial velocity, but in the opposite direction. Therefore, v = -v0. Plugging in the values for a (9.8 m/s2) and t (1.56 s), we get:
-v0 = v0 + (9.8 m/s2)(1.56 s)
Solving for v0, we get:
v0 = (9.8 m/s2)(1.56 s)/2
v0 = 7.644 m/s
The initial velocity of the ball is 7.644 m/s in the upward direction.
F) The maximum height that the ball reaches can be found using the equation h = v0t + (1/2)at2, where h is the height, v0 is the initial velocity, t is the time, and a is the acceleration.
Plugging in the values for v0 (7.644 m/s), t (1.56 s/2 = 0.78 s), and a (9.8 m/s2), we get:
h = (7.644 m/s)(0.78 s) + (1/2)(9.8 m/s2)(0.78 s)2
h = 2.977 m
The maximum height that the ball reaches is 2.977 m.
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A car approaches a stop sign going 5 m/s before it comes to a stop. If it does this in 4 seconds, what was its acceleration?
Answer:
[tex]a = - \frac{5}{4} \: m{s}^{ - 2}[/tex]
Explanation:
a = acceleration (m/s²)
v = final velocity (m/s) = 0
u = initial velocity (m/s) = 5
t = time (s) = 4
[tex]a = \frac{v - u}{t} \\ a = \frac{(0 - 5)}{4} \\ a = - \frac{5}{4} [/tex]
What is the unknown mass of the second box shown in the collision below?
Before collision: v= 6 m/s, 7kg
stationary: m
After collision: v= 2 m/s, 7 kg, m
Answer:
We can use conservation of momentum to solve for the unknown mass of the second box. The total momentum before the collision is equal to the total momentum after the collision.
Before collision:
p = m1v1 = (7 kg)(6 m/s) = 42 kg m/s
After collision:
p = m1v1 + m2v2 = (7 kg)(2 m/s) + m(2 m/s)
Setting the two expressions for momentum equal to each other, we can solve for m2:
42 kg m/s = (7 kg)(2 m/s) + m(2 m/s)
42 kg m/s = 14 kg m/s + 2m kg m/s
28 kg m/s = 2m kg m/s
m = 14 kg
Therefore, the mass of the second box is 14 kg
A man fires a shot while in a deep canyon with parallel cliff walls. The temperature in the canyon is 27.0°C. If the echo from one wall is heard in 1.50 seconds, and the echo from the other wall is heard 1.00 second later, how wide is the canyon?
The canyon is approximately 258 meters wide.
What is the width of the canyon?The speed of sound in air depends on the temperature, and we can use the formula:
v = 331.4 + 0.6T
where;
v is the speed of sound in meters per second, and T is the temperature in Celsius. At a temperature of 27.0°C,The speed of sound is:
v = 331.4 + 0.6(27.0)
v = 343.8 m/s
Let's call the distance to one wall "d". When the man fires the shot, the sound travels to the wall, reflects, and returns to the man.
The total distance traveled is 2d. The time it takes for the sound to travel this distance is:
t = 2d/v
Similarly, for the other wall, the total distance traveled is 2(D - d)
where;
D is the width of the canyon.The time it takes for the sound to travel this distance is:
t + 1.00 = 2(D - d)/v
We can solve these equations for d and D:
d = vt/2
D = (v/2)(t + 1.00)
Substituting in the values we have:
d = (343.8 m/s)(1.50 s)/2 = 257.85 m
D = (343.8 m/s/2)(1.50 s + 1.00 s) = 515.70 m
Therefore, the width of the canyon is:
D - d = 515.70 m - 257.85 m = 257.85 m
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A rectangular block with dimensions 6.0 cm x 8.0 cm x 12.0 cm is made of aluminium of density 2700 kg m¹. Find the maximum pressure it can exert when placed on one of its faces on a horizontal surface.
The maximum pressure the rectangular block can exert on a surface when placed on one of its faces on a horizontal surface is 317.71 Pa.
What is the maximum pressure of the block?The maximum pressure the rectangular block can exert on a surface occurs when it is placed on one of its smallest faces.
To calculate the maximum pressure, we need to find the weight of the block, which is equal to its mass multiplied by the acceleration due to gravity. The mass of the block can be calculated as follows:
mass = density x volume
The volume of the block is:
volume = length x width x height
volume = 6.0 cm x 8.0 cm x 12.0 cm
volume = 576 cm³
Converting to meters:
volume = 0.006 m x 0.008 m x 0.12 m
volume = 5.76 x 10^-5 m³
Therefore, the mass of the block is:
mass = density x volume
mass = 2700 kg/m³ x 5.76 x 10^-5 m³
mass = 0.1555 kg
The weight of the block is:
weight = mass x acceleration due to gravity
weight = 0.1555 kg x 9.81 m/s²
weight = 1.526 N
When the block is placed on one of its smallest faces, the area of contact with the surface is:
area = length x width
area = 6.0 cm x 8.0 cm
area = 48 cm²
Converting to meters:
area = 0.06 m x 0.08 m
area = 0.0048 m²
Therefore, the maximum pressure the block can exert on the surface is:
pressure = weight / area
pressure = 1.526 N / 0.0048 m²
pressure = 317.71 Pa
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Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint of the distance is marked to make it easier to see how the locations compare. Assume the spaceship has the same mass throughout the trip (that is, it is not burning any fuel). Rank the five positions of the spaceship from left to right based on the strength of the gravitational force that Earth exerts on the spaceship, from strongest to weakest. The following diagrams are the same as those from Part A. This time, rank the five positions of the spaceship from left to right based on the strength of the gravitational force that the Moon exerts on the spaceship, from strongest to weakest. The following diagrams show five pairs of asteroids, labeled with their relative masses (M) and distances (d) between them. For example, an asteroid with M=2 has twice the mass of one with M=1 and a distance of d= 2 is twice as large as a distance of d = 1. Rank each pair from left to right based on the strength of the gravitational force attracting the asteroids to each other, from strongest to weakest.
A: The gravity between Earth and the spaceship decreases, B: Gravity obeys the inverse square law, C: d=2; m=2; m=2, d=1; m=1; m=2, d=1; m=1; m=1, d=2; m=1; m=2, d=2; m=1; m=1.
What is the gravitational equivalent of the inverse square law?According to Newton's inverse square law, the force of gravity acting between any two objects is inversely equal to the square of the distance between their centres. Altering the separation distance (d) results in a modification in the force of gravity acting between the objects.
How did Newton arrive at the inverse square law?The third law of planetary motion, which states that "the square of the time taken for a planet to orbit the sun is directly proportional to the cube of the semi-major axis of the orbit," is where Newton actually got the idea for his inverse square law.
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a proton is near a charged plate and feels a force of 4.5x10^-16 N if the oroton is 1.3x1^-6 m away from the charged plate what is the charge of the plate
The wires in a piano vibrate, but fractions of the wire also vibrate at different frequencies than the whole wire. What is the term
for the softer notes produced by these different frequencies? (1 point)
O interference
O pitch
O resonance
Oharmonics
the answer to the question is pitch
A sample of helium gas occupies 5 liters at 22°C and a gauge pressure of 200,000 pascals. What is the mass of the gas? (The gas constant for helium is 2.08 x 103 J/kg-K.)
A sample of helium gas occupies 5 liters at 22°C and a gauge pressure of 200,000 pascals. The mass of the gas is 0.16 gm. It is calculated using ideal gas equation.
Define gauge pressure.
When compared to atmospheric pressure, gauge pressure is the pressure that is higher; it is positive for pressures over atmospheric pressure and negative for pressures below atmospheric pressure. Every fluid that is not contained experiences additional pressure due to the atmospheric pressure.
Gauge pressure, which is equal to the difference between absolute and atmospheric pressure and is zero-referenced against ambient air pressure, is the additional pressure in any system relative to atmospheric pressure. The absence of the negative sign serves to distinguish the negative pressure and is typically ignored.
The word "vacuum" can be given a value, and the gauge is referred to as a vacuum gauge. When internal pressure exceeds atmospheric pressure in a given area, the phrase gauge pressure is employed.
Using the formula:
PV = nRT
substituting the values and solving for n:
n = 0.04 moles
mass = 0.04 × 4 = 0.16g
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till what time the balloon expands when the pressure of outside air is greater than than the inside pressure or equal?
Calculate the magnitude of the gravitational force exerted by Venus on a 65 kg human standing on the surface of Venus. (The mass of Venus is 4.9x1024 kg and its radius is 6.1x106 m.) N Calculate the magnitude of the gravitational force exerted by the human on Venus. N For comparison, calculate the approximate magnitude of the gravitational force of this human on a similar human who is standing 3.5 meters away. N What approximations or simplifying assumptions must you make in these calculations? (Note: Some of these choices are false because they are wrong physics!) Treat the humans as though they were points or uniform-density spheres. Ignore the effects of the Sun, which alters the gravitational force that one object exerts on another. Treat Venus as though it were spherically symmetric. Use the same gravitational constant in (a) and (b) despite its dependence on the size of the masses.
To calculate the magnitude of the gravitational force exerted by Venus on a 65 kg human standing on the surface of Venus, we use Newton's Law of Universal Gravitation and it will be 6.3 x 10-6 N
This states that the gravitational force (F) between two masses (m1 and m2) separated by a distance (r) is given by the equation:
F = G * (m1 * m2) / r2
Where G is the gravitational constant, which is 6.67 x 10-11 Nm2/kg2. Using this equation and the given values, we have:
F = 6.67 x 10-11 * (4.9 x 1024 * 65) / (6.1 x 106)2
F = 3.3 x 105 N
To calculate the magnitude of the gravitational force exerted by the human on Venus, we use the same equation as before, but with the masses of the human and Venus reversed. This gives:
F = 6.67 x 10-11 * (65 * 4.9 x 1024) / (6.1 x 106)2
F = 4.0 x 10-6 N
For comparison, the approximate magnitude of the gravitational force of the human on a similar human who is standing 3.5 meters away can be calculated using the same equation. The masses of both humans are the same, so we have:
F = 6.67 x 10-11 * (65 * 65) / (3.5)2
F = 6.3 x 10-6 N
To calculate the gravitational force in both cases, we had to make the following approximations or simplifying assumptions:
Treat the humans as though they were points or uniform-density spheres.
Ignore the effects of the Sun, which alters the gravitational force that one object exerts on another
And Treat Venus as though it were spherically symmetric.
We also had to use the same gravitational constant in (a) and (b) despite its dependence on the size of the masses.
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The speed of a wave is equal to the frequency multiplied by... what?
Answer: The wavelength
Explanation:
The speed of a wave is calculated as the product of the frequency times the wavelength.
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The reactants of a chemical equation have 1 S atom and 4 O atoms. Which set of atoms must also be founding the equations products so that the equation models the law of conservation of mass
A. 1 S and 4 O
B. 1 S and 1 O
C. 4 S and 1 O
D. 4 S and 4O
The set of atoms must also be founding the equations products so that the equation models the law of conservation of mass is option (A) 1 S and 4 O
The law of conservation of mass is a fundamental principle in chemistry that states that the mass of the reactants in a chemical reaction must be equal to the mass of the products. This principle is based on the fact that atoms cannot be created or destroyed in a chemical reaction, and therefore, the number of atoms of each element on both sides of the equation must be the same.
In order to balance a chemical equation and satisfy the law of conservation of mass, it is necessary to adjust the coefficients of the reactants and products to ensure that the number of atoms of each element is equal on both sides of the equation.
Therefore, the correct option is (A) 1 S and 4 O
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Which property of the Sun most affects the
strength of gravitational attraction between
the Sun and Earth?
A mass
B radius
C shape
D temperature
Answer:
mass
Explanation:
A fan turns a rate 900 rpm. Find the angular speed of any point on one of the fan blades and find the tangential speed of the tip of a blade if the distance from the center to the tip is 20 cm.
a. 15 rad/s & 3 m/s
b. 15 rad/s & 30 m/s
c. 94.2 rad/s & 18.8 m/s
d. 900 rad/s & 180 m/s
At 900 rpm, a fan spins. angular speed of the any point on a fan blade and tangential speed of a blade's tip if there is a 20 cm gap between the centre and the tip. 9.4 rads and 18.8 m/s.
Correct option is, C.
What is the formula for angular velocity?A amount or angle rotated (and angular displacement) by a rotating body in a given length of time is known as the angular velocity of the body. The symbol for it is omega (). The formula for angular velocity is rad/s = dtd. Radian per second serves as its SI unit.
What is the SI unit for angular momentum?The product of a moment of inertia (I) as well as the angular velocity () of the an object in rotation is the angular momentum. A vector quantity is angular momentum. Kg. m2 is the SI unit for angular momentum.
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Compute the resistance of a 90-cm length of copper wire with a 0. 020 -cm^2 cross-sectional area. ( = 1. 8 x 10-6 ohm-cm)
R = ______ ohms
The resistance of a 90-cm length of copper wire with a 0. 020 -cm^2 cross-sectional area is 81 ohms.
The resistance of a wire is given by:
R = (rho * L) / A
where rho is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
In this case, we are given that the length of the copper wire is 90 cm, the cross-sectional area is 0.020 cm^2, and the resistivity of copper is 1.8 x 10^-6 ohm-cm.
First, we need to convert the units of length and cross-sectional area to meters and square meters, respectively:
L = 90 cm = 0.9 m
A = 0.020 cm^2 = 2 x 10^-5 m^2
Now, we can substitute these values into the formula for resistance:
R = (rho * L) / A
R = (1.8 x 10^-6 ohm-cm * 0.9 m) / 2 x 10^-5 m^2
R = 81 ohms
Therefore, the resistance of the copper wire is 81 ohms.
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Simple Pendulum Shows Uniform or Non uniform Motion. Explain?
The speed of the bob in a simple pendulum changes every time from A to B. So, its motion can't be a uniform motion. Therefore, the motion of the bob in a simple pendulum is a non-uniform motion.
A simple pendulum is a weight (or bob) suspended from a string or rod, which is free to swing back and forth under the influence of gravity. It is an idealized model used to study the behavior of more complex pendulum systems and has been studied extensively in physics due to its simplicity and ubiquity in everyday life.
The motion of a simple pendulum can be described by a number of variables, including the length of the string, the mass of the bob, the angle at which the pendulum is released, and the amplitude and period of the pendulum's oscillations.
In the absence of friction, the motion of a simple pendulum is governed by the laws of conservation of energy and conservation of momentum.
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Calculate the pressure exerted on the floor when an elephant who weighs 6000
lb stands on one foot which has an area of 20 in².
Answer:
Explanation:
To calculate the pressure exerted on the floor, we need to use the formula:
pressure = force / area
where force is the weight of the elephant and area is the area of the foot in contact with the floor.
First, we need to convert the weight of the elephant from pounds to pounds-force (lbf), which is the force exerted by a mass due to gravity. We can do this by multiplying the weight in pounds by the acceleration due to gravity, which is approximately 32.2 ft/s²:
force = weight x acceleration due to gravity
= 6000 lb x 32.2 ft/s²
= 193200 lbf
Now we can calculate the pressure:
pressure = force / area
= 193200 lbf / 20 in²
≈ 9660 psi
Therefore, the pressure exerted on the floor by the elephant standing on one foot is approximately 9660 pounds per square inch (psi). This is a very high pressure and could cause damage to some types of flooring.
3. show your calculations for the following quantities for 1.5 v. include formula and show all your work.
The calculation for the quantity for 1.5 V is simple: 1.5V = 1.5 x 1 = 1.5. The formula for this calculation is V = Voltage x 1, where V is the quantity. The calculation for 1.5V is shown below:
1.5V = 1.5 x 1 = 1.5, By multiplying 1.5 with 1, we get 1.5. This is the calculation for 1.5V.
To understand the calculation of 1.5V better, it is important to first understand the basic unit of electricity - volts. Volts measure the force of electricity in a circuit. It is the amount of electricity that flows through the circuit and is measured in terms of voltage. The higher the voltage, the more electricity is available in the circuit.
To calculate the quantity of electricity for 1.5V, we need to multiply 1.5 with 1. This is the calculation for 1.5V. By multiplying 1.5 with 1, we get 1.5, which is the quantity of electricity for 1.5V.
In conclusion, the calculation for the quantity of electricity for 1.5V is 1.5 x 1 = 1.5. This calculation can be represented using the formula V = Voltage x 1, where V is the quantity.
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what is the speed of an object with a mass of 192 kg when it hits the ground if it is dropped from a height of 13 m
Answer:
V=15.96M/S OR ～16M/SExplanation:
SINCE IT IS DROPPED, INITIAL VELOCITY WILL BE ZEROU = 0
V = ?
S = 13M
ACCELERATION DUE TO GRAVITYA = 9.8m/s²
USING KINEMATIC FORMULA[tex]v {}^{2} = u {}^{2} + 2as[/tex]
[tex]v { }^{2} = 0 + 2(9.8)(13) \\ v {}^{2} = 254.8 \\ v = 15.96m/s[/tex]
A test charge Is Introduced Into the electric field of a charge. It feels a force of 2F where the electric field is 2E. What
would the force on the test charge be where the electric field is 8E?
When the electric field is 8E, the force acting on the test charge would be 8F.
What power does an electric field exert on a test charge?The quantity of force applied to a test charge per unit of charge is known as the electric field strength (E). (q). As a result, E = F / q. According to Coulomb's rule, there are a number of factors that affect the electric force (F).
In line with the formula:
Force (F)=Charge x Electric Field
We can use this knowledge to determine the charge on the test charge if it experiences a force of 2F where the electric field is 2E:
2F = q × 2E
q = 2F / 2E
q = F / E
The force acting on the test charge where the electric field is 8E can now be determined using the charge number mentioned above:
Force (F') = Charge (q) x Electric Field (E')
F' = q × E'
F' = (F / E) × 8E
F' = 8F.
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