false. The general momentum equation may be used to evaluate both elastic and inelastic collisions, but it cannot tell whether a collision is elastic or inelastic on its own.
The quantity of kinetic energy preserved during the impact determines the definition of elastic and inelastic collisions. Kinetic energy is conserved in an elastic collision, which means that the total kinetic energy before and after the impact stays constant. Some kinetic energy is lost in an inelastic collision, and the total kinetic energy after the collision is less than the total kinetic energy before the contact. To evaluate whether a collision is elastic or inelastic, one must examine the amount of kinetic energy preserved during the collision, in addition to applying the general momentum equation.
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a hollow spherical shell has mass 7.90 kg and radius 0.230 m . it is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.895 rad/s2 . part a what is the kinetic energy of the shell after it has turned through 5.00 rev ?
The hollow spherical shell has a mass of 7.90 kg and a radius of 0.230 m. It initially rests and then rotates with a constant acceleration of 0.895 rad/s2 around a stationary axis that lies along a diameter. The kinetic energy of the hollow spherical shell is 2.12 J.
The first step is to calculate the angular displacement using the formulaθ = n × 2π = 5.00 rev × 2π/rev = 31.4 rad[The angular displacement here is a positive value, as the spherical shell is rotating in a counterclockwise direction]. The next step is to calculate the angular velocity after 5.00 rev, using the formula
ωf = ωi + αt, where ωi = 0 [initial angular velocity]α = 0.895 rad/s2 [angular acceleration n]t = 2.22 s [time taken to complete 5.00 revolutions]Therefore,ωf = 0 + 0.895 × 2.22 = 1.987 rad/s Kinetic energy of the shell, K = 1/2 I ω²where I is the moment of inertia of the shell.
The moment of inertia of a hollow spherical shell is given by
I = 2/3 M R² where M is the mass of the shell and R is the radius of the shellSubstituting values, K = 1/2 × 2/3 × 7.90 × (0.230)² × (1.987)²= 2.12 J [to 2 significant figures]
Therefore, the kinetic energy of the hollow spherical shell is 2.12 J after it has turned through 5.00 revolutions.
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which one of the following quantities remains constant for a given lc circuit? group of answer choices the energy dissipated in the circuit. the sum of the energy stored in the capacitor and that in the inductor. the energy stored in the inductor. the energy stored in the current flowing in the circuit. the energy stored in the capacitor.
For the given LC circuit (inductor-capacitor) , the sum of the energy stored in the capacitor and that in the inductor remains constant.
Energy transferred between the inductor and capacitor. So the total energy oscillating between the two components at a resonant frequency.
The energy stored in the capacitor and inductor individually varies over time as the energy oscillates between them.
When voltage changes the energy stored in the capacitor also changes. This is caused by the change in current flowing.
So we can say both the energy stored in the capacitor and the energy stored in the inductor remains constant in a given LC circuit.
The energy dissipated in the circuit can vary over time. Like that the energy stored in the current flowing in the circuit and the energy stored in the capacitor can also vary over time in an LC circuit.
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a 80 kg man lying on a surface of negligible friction shoves a 53 g stone away from himself, giving it a speed of 4.6 m/s. what speed does the man acquire as a result?
When an 80 kg man moving at 120.75 m/s pushes a 53 g stone away from him while lying on a surface with little friction, the stone moves at a speed of 4.6 m/s.
Let the man acquire the speed v in the opposite direction.Let the momentum be conserved here.The momentum of the stone before the push is: p₁ = 0The momentum of the stone after the push is: p₂ = m × vWhere m is the mass of the stoneThe impulse is given as: J = p₂ - p₁Now, we know that the impulse (J) = Force (F) × time (t).
We also know that force is mass × acceleration. Therefore, the impulse can be written as: J = m × a × tUsing these equations we can solve for the acceleration (a).a = J/(m × t)Now, the acceleration is the same for the man and the stone, but the masses are different.
Therefore, the man acquires a speed v that is much smaller than the velocity of the stone. Substituting the given values we get,a = (m₂v₂ - m₁v₁)/(m₂t₂) = (0.053 × 4.6)/(80 × t) = 0.00109/t m/s². After equating the forces acting on both the stone and the man, we have;Fman = - Fstone.
This is because the man's speed is in the opposite direction to the stone.Let u be the initial speed of the man before he shoves the stone away from himself.
Using the momentum formula, m1u1 + m2u2 = m1v1 + m2v2.The mass of the stone is 0.053 kg while the man's mass is 80 kg.So,80u + 0.053 × 0 = 80v + 0.053 × 4.6v = (80u) / 0.053+4.6v = (80u) / 0.053v = 120.75u.
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what is the energy density (energy per mass) of butter and similar foods? group of answer choices 4.5 mj/kg 4500 mj/kg 45 mj/kg 450 mj/kg 0.45 mj/kg
The energy density of butter and similar foods is 4.5 MJ/kg. This value represents the amount of energy stored in a given mass of food and is related to the food's calorie content.
The energy density (energy per mass) of butter and similar foods can be measured in megajoules per kilogram (MJ/kg). Energy density is important for understanding the amount of energy stored in a given mass of food, which is related to its calorie content.
1. Determine the macronutrient content of the food (i.e., grams of fat, carbohydrates, and protein per serving).
2. Multiply the grams of fat by 9 kcal/g, as fat provides 9 kcal of energy per gram.
3. Multiply the grams of carbohydrates and protein by 4 kcal/g, as both carbohydrates and proteins provide 4 kcal of energy per gram.
4. Add the energy content from each macronutrient to get the total energy content of the food.
5. Divide the total energy content by the mass of the food (in kg) to obtain the energy density (in MJ/kg).
Among the given answer choices, the correct energy density for butter and similar foods is 4.5 MJ/kg. This value is equivalent to 1,000 kcal/kg, as there are approximately 0.004184 MJ in 1 kcal. Butter and similar high-fat foods have high energy densities due to their high fat content, which provides more energy per gram compared to carbohydrates and proteins.
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a ball of mass 5.0 kg is lifted off the floor a distance of 1.7m. what is the change in gravitational potential energy of th eball?
The change in gravitational potential energy of the ball is 83.3 J.
When a ball of mass 5.0 kg is lifted off the floor a distance of 1.7m, what is the change in gravitational potential energy of the ball?The change in gravitational potential energy of the ball is determined by the formula given below;
ΔPEg=mgh
Where;ΔPEg = Change in gravitational potential energy.m = Mass of the object.
g = Acceleration due to gravity.
h = Height of the object.The mass of the ball is given as 5.0 kg,
and the height it was lifted is 1.7 m.
Acceleration due to gravity, g, is 9.8 m/s2.
Substituting these values in the above formula gives the change in gravitational potential energy of the ball.
ΔPEg= mgh
= 5.0 kg × 9.8 m/s2 × 1.7
m= 83.3 J
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A 3kg crab was moving at 1 m/s in the shore before the ride pushed him for 5 seconds. If his final speed was 3 m/s, what force did the tide push him with?
Answer:
[tex]1.2\; {\rm N}[/tex], assuming that all other forces on this crab were balanced.
Explanation:
The impulse [tex]J[/tex] on an object is equal to the change in momentum [tex]\Delta p[/tex]. In other words:
[tex]J = \Delta p[/tex].
If the mass [tex]m[/tex] of the object stays the same (as in the case of this question), the change in momentum can be rewritten as:
[tex]J = \Delta p = m\, \Delta v[/tex], where [tex]\Delta v[/tex] is the change in velocity.
Impulse is also equal to the net force on the object [tex]F_{\text{net}}[/tex] times the duration [tex]\Delta t[/tex] over which the force is applied:
[tex]J = F_{\text{net}}\, \Delta t[/tex].
Equate the two expressions for [tex]J[/tex] to obtain:
[tex]F_{\text{net}}\, \Delta t = m\, \Delta v[/tex].
In this question:
[tex]\Delta t = 5\; {\rm s}[/tex] is the duration over which the force was applied,[tex]m = 3\; {\rm kg}[/tex] is the mass of the crab, and[tex]\Delta v = (3 - 1)\; {\rm m\cdot s^{-1}} = 2\; {\rm m\cdot s^{-1}}[/tex] is the change in the velocity of the crab.Rearrange [tex]F_{\text{net}}\, \Delta t = m\, \Delta v[/tex] and solve for the net force [tex]F_{\text{net}}[/tex]:
[tex]\begin{aligned}F_{\text{net}} &= \frac{m\, \Delta v}{\Delta t} \\ &= \frac{(3\; {\rm kg})\, (2\; {\rm m\cdot s^{-1}})}{5\; {\rm s}} \\ &= 1.2\; {\rm kg \cdot m\cdot s^{-2}} \\ &= 1.2\; {\rm N}\end{aligned}[/tex].
Assuming that all other forces on this crab are balanced, the net force on the crab would be equal to the force from the tide. Hence, the tide would have pushed the crab with a force of [tex]1.2\; {\rm N}[/tex].
in a drum-buffer-rope system, the lot size that moves from one work center to another for additional processing is a(n)
In a drum-buffer-rope system, the lot size that moves from one work center to another for additional processing is a "batch" size.
The drum-buffer-rope (DBR) system is a manufacturing scheduling method that aims to maximize throughput while minimizing inventory and operating expenses. In this system, the "drum" represents the bottleneck work center that determines the production rate for the entire system, and the "buffer" is a quantity of inventory strategically placed before the bottleneck to prevent downtime.
The "rope" refers to the mechanism used to synchronize production with demand, typically through a pull-based system that only allows material to be released to the next work center when the previous work center has completed its processing. This approach ensures that production is driven by actual demand and that inventory is minimized throughout the system, resulting in improved efficiency and profitability.
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--The complete question is, in a drum-buffer-rope system, the lot size that moves from one work center to another for additional processing is ______.--
Which of the following best describes statistics?
1. a way to organize data
2. a way to analyze data
3. all are correct
4. a way to interpret data
The following best describes statistics : 2.) a way to analyze data.
What is statistics?Statistics can be described as a way to analyze data. It includes the collection, organization, analysis, interpretation and presentation of data. Statistics deals with gathering, presenting and also arranging of information to make any decision.
Study and manipulation of data, including ways to gather, review, analyze, and draw conclusions from data is called statistics and two major areas of statistics are descriptive and inferential statistics.
Statistics are important as they help people make informed decisions. Governments, organizations and businesses collect statistics that helps them to track the progress, measure performance and then analyze problems.
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a 63 kg k g person starts traveling from rest down a waterslide 4.0 m m above the ground. at the bottom of the waterslide, it then curves upwards by 1.0 m m above the ground such that the person is consequently launched into the air. ignoring friction, how fast is the person moving upon leaving the waterslide? express your answer with the appropriate units.
Person is moving at most at a speed of 11.8 m/s
We can use the principle of conservation of energy. In this scenario, the person starts at a height of 4.0 m and ends at a height of 5.0 m. Using the formula for gravitational potential energy, we can calculate that the initial potential energy is
[tex]63 kg * 9.81 m/s^2 * 4.0 m = 2474.04 J.[/tex]
At the top of the curve, all of this energy is converted into potential energy again, so the kinetic energy is zero.
[tex]63 kg * 9.81 m/s^2 * 5.0 m = 3085.05 J.[/tex]
Equating these energies, we get[tex]1/2 mv^2 = 3085.05 J[/tex],
where m is the mass of the person and v is the velocity. Solving for v, we get [tex]v = \sqrt{(2 * 3085.05 J / 63 kg)} = 11.8 m/s[/tex].
Therefore, the person is moving at a speed of 11.8 m/s upon leaving the waterslide.
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starting from rest, a solid sphere rolls without slipping down an incline plane. at the bottom of the incline, what does the angular velocity of the sphere depend upon? check all that apply.
Angular velocity depends upon the radius of the sphere and the height of the incline when a solid sphere rolls without slipping down an incline plane. The correct options are b and c.
A sphere is a three-dimensional shape that is circular and completely round, like a ball or globe. Solid spheres, also known as balls, are objects with a radius and a center point that are fully enclosed by a curved surface in three dimensions.
The angular velocity of a solid sphere rolling down an incline plane without sliding depends on the radius of the sphere and the height of the inclination.
The angular velocity is unaffected by the length of the slope or the mass of the sphere. This is due to the fact that the angular velocity is governed by the sphere's moment of inertia and rotational kinetic energy, both of which are dictated by the sphere's form (radius) and the height of the incline.
Therefore, options are b and c are correct.
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The probable question may be:
Starting from rest, a solid sphere rolls without slipping down an incline plane. at the bottom of the incline, what does the angular velocity of the sphere depend upon? check all that apply.
a. The angular velocity depends upon the length of the incline b. The angular velocity depends upon the radius of the sphere c. The angular velocity depends upon the height of the incline d. The angular velocity depends upon the mass of the sphere
The resistance produced by a current of 120 amps from a 6V battery is??
[tex]Answer[/tex]
To find the resistance produced by a current of 120 amps from a 6V battery, we can use Ohm's law, which states that the voltage (V) across a resistor is equal to the current (I) flowing through the resistor multiplied by the resistance (R) of the resistor:
V = IR
In this case, we know that the current is 120 amps and the voltage is 6V, so we can rearrange the equation to solve for the resistance:
R = V/I
Substituting the given values, we get:
R = 6V / 120A = 0.05 ohms
Therefore, the resistance produced by a current of 120 amps from a 6V battery is 0.05 ohms.
a ball traveling in a circle with a constant speed of 15 m/s has a centripetal acceleration of 20 m/s2. what is the radius of the circle?
The radius of the circle can be calculated using the formula; Centripetal acceleration = v^2/r, where v is the speed and r is the radius of the circle.
Substitute the given values; Centripetal acceleration = 20 m/s^2Speed = 15 m/s Using the formula above, we have;20 = (15)^2/rr = (15)^2/20r = 11.25mTherefore, the radius of the circle is 11.25 m.
The radius of the circle can be calculated using the formula for centripetal acceleration: a_c = v^2 / r, where a_c is the centripetal acceleration, v is the speed, and r is the radius. In this case, a_c = 20 m/s^2 and v = 15 m/s. Rearranging the formula to find r, we get:
r = v^2 / a_c = (15 m/s)^2 / (20 m/s^2) = 225 m^2 / 20 m/s^2 = 11.25 m
The radius of the circle is 11.25 meters.
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what interferences can be subtracted out of an ir spectrum due to the usage of a double beam spectrometer with a reference cell?
Explanation: In a double beam spectrometer with a reference cell, interferences due to the light source, the environment, and the sample holder can be subtracted out of an air spectrum.
The reference cell is filled with a gas that is transparent and has no absorption bands in the spectral range of interest. The reference cell and the sample cell are alternatively placed in the path of the light source. The detector records the intensity of the light passing through both cells, which is then compared to the intensity of the light passing through the sample cell only.
By using the reference cell as a baseline, any interferences due to the light source or the environment (e.g., variations in temperature, pressure, or humidity) can be subtracted out, as these will affect the intensity of the light passing through both cells equally. Similarly, any interferences due to the sample holder or any contaminants in the sample can be subtracted out, as these will affect the intensity of the light passing through the sample cell only.
The result is a spectrum that reflects only the absorption or transmission of the sample itself, without interference from other sources.
When using a double beam spectrometer with a reference cell, several interferences can be subtracted out of an IR spectrum which are: Background absorption, Stray light, Sample cell absorption and Instrumental noise.
What is a double beam spectrometer?A spectrometer is an instrument that measures the light intensity and the wavelength of light. The double beam spectrometer works on the principle of the single beam spectrometer but with two beams of light. The use of a double beam spectrometer with a reference cell enables you to measure the absorption of infrared radiation at different wavelengths due to a specific chemical compound.
Interferences that can be subtracted out of an IR spectrum due to the usage of a double beam spectrometer with a reference cell:
1. Background absorption: Background absorption occurs when an IR beam from the source is absorbed by the atmosphere or by the instrument components. This absorption can be reduced by subtracting the IR spectrum of the reference material from the spectrum of the sample.
2. Stray light: Stray light is a problem that occurs when the IR beam from the source is not completely focused on the sample. This leads to a low intensity of the IR beam reaching the detector. It can be reduced by the use of appropriate filters.
3. Sample cell absorption: This is the absorption of the sample cell material that contributes to the IR spectrum. This can be corrected by subtracting the spectrum of the empty reference cell from the spectrum of the sample.
4. Instrumental noise: This is the background noise in the IR spectrum that is due to the instrument. It can be reduced by using appropriate signal processing techniques.
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A) How much gravitational potential energy must a 3120-kg satellite acquire in order to attain a geosynchronous orbit?
B) How much kinetic energy must it gain? Note that because of the rotation of the Earth on its axis, the satellite had a velocity of 456 m/s relative to the center of the Earth just before launch.
The kinetic energy required for the satellite to attain a geosynchronous orbit is approximately 3.00 × 108 J.
When answering questions on Brainly, a question-answering bot should always be factually accurate, professional, and friendly, be concise and not provide extraneous amounts of detail, and use the following terms provided in the student question while also ignoring any typos or irrelevant parts of the question.
In response to the student question, we will calculate the gravitational potential energy and kinetic energy required for a 3120-kg satellite to attain a geosynchronous orbit.
Gravitational potential energyTo calculate the gravitational potential energy required to attain a geosynchronous orbit, we use the formula:U = -G (Mm/r)where U is the gravitational potential energy, G is the universal gravitational constant (6.67 × 10-11 N m2/kg2), M is the mass of the Earth (5.98 × 1024 kg),
m is the mass of the satellite (3120 kg), and r is the radius of the geosynchronous orbit (42,164 km).Substituting these values into the equation:U = -6.67 × 10-11 × 5.98 × 1024 × 3120 / 42,164 × 103U = -9.84 × 1010 JSo the gravitational potential energy required for the satellite to attain a geosynchronous orbit is approximately 9.84 × 1010 J.
Kinetic energyTo calculate the kinetic energy required to attain a geosynchronous orbit, we use the formula:KE = (1/2)mv2where KE is the kinetic energy,
m is the mass of the satellite (3120 kg), and v is the velocity of the satellite just before launch (456 m/s).Substituting these values into the equation:KE = (1/2) × 3120 × (456)2KE = 3.00 × 108 J
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in what fundamental respect does electromagnetism break away from the form of materialism associated with the physics of newton and democritus? group of answer choices the electromagnetic force can be felt across a vacuum. newton had thought that the only force in the universe was gravity. the electromagnetic field is physically real, even though it is not made of atoms. nonsense--electromagnetism agrees quite well with newtonian materialism. electromagnetism implies that the future cannot be precisely determined.
Electromagnetism fundamentally breaks away from the materialism associated with the physics of Newton and Democritus in that the electromagnetic field is physically real, even though it is not made of atoms.
This distinction is significant because both Newtonian physics and Democritus atomism rely on the idea that matter, made up of atoms or other particles, is the primary building block of the universe. Newtonian physics, based on the concept of gravity as the only force in the universe, focuses on the interaction of massive objects and their motion.
Democritus, an ancient Greek philosopher, proposed that the universe was made up of indivisible particles called atoms, which form various materials through their combinations. On the other hand, electromagnetism introduces the concept of electric and magnetic fields, which are not made of material particles or atoms.
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if a diver were to descend to 10 m with air in her lungs without breathing in or out, what would be the approximate new value for the volume of the air in her lungs? treat the lung as 6 l of ideal gas.
Answer:
Its C
Explanation:
I took the test
As the volume of air in the lungs of the diver will reduce when she descends to 10m with air in her lungs without breathing in or out, the approximate new value will be 17.56 L, calculated using Boyle's law.
Boyle's law states that the pressure exerted by a given mass of gas is inversely proportional to its volume at a constant temperature. The mathematical expression for Boyle's law is: P1V1 = P2V2where, P1 and V1 are the initial pressure and volume of the gas, respectively, and P2 and V2 are the final pressure and volume of the gas, respectively.
Let V1 be the initial volume of air in the lungs of the diver, which is 6L. Let P1 be the atmospheric pressure at the surface of the water, which is 1 atm. Let P2 be the pressure at 10 m depth, which can be calculated using the hydrostatic equation: P2 = P1 + ρghwhere, ρ is the density of water, g is the acceleration due to gravity, and h is the depth of the water.
The density of water at standard temperature and pressure is 1000 kg/m3. Therefore, ρgh = 1000 kg/m3 × 9.81 m/s2 × 10 m = 98,100 Pa. Hence, P2 = 1 atm + 98,100 Pa = 2.05 atm. Substituting the values in Boyle's law equation, we get:1 atm × 6 L = 2.05 atm × V2. Therefore, V2 = (1 atm × 6 L) / 2.05 atm = 17.56 L. Approximately, the new value for the volume of air in the lungs of the diver would be 17.56 L.
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The mean, median and mode are examples of what type of statistics?
1. experimental
2. inferential
3. descriptive
4. correlatoinal
Answer:
3. descriptive
Explanation:
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the amplitude of a 3.00-kg object in simple harmonic motion is 6.00 m. the maximum acceleration of the object is 5.00 m/s2. what is the period of simple harmonic motion?
The period of simple harmonic motion for this 3.00-kg object is approximately 6.87 seconds.
To find the period of simple harmonic motion for a 3.00-kg object with an amplitude of 6.00 m and a maximum acceleration of 5.00 m/s², we first need to find the angular frequency (ω).
We know the formula for the maximum acceleration in simple harmonic motion is given by:
amax = ω² * A
where amax is the maximum acceleration, A is the amplitude, and ω is the angular frequency.
Rearranging the formula to solve for ω, we get:
ω = sqrt(amax / A)
Plugging in the given values:
ω = sqrt(5.00 m/s² / 6.00 m)
ω ≈ 0.912 m^(-1/2) s^(-1)
Now that we have the angular frequency, we can find the period (T) using the relationship between angular frequency and period:
ω = 2π / T
Rearranging the formula to solve for T, we get:
T = 2π / ω
Plugging in the value of ω we found earlier:
T ≈ 2π / 0.912 m^(-1/2) s^(-1)
T ≈ 6.87 s
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suppose you moved two objects farther apart. how would this affect the force of gravity between those objects?
The moving objects farther apart will result in a weaker gravitational attraction between them.
When two objects are moved farther apart, the force of gravity between them decreases. This relationship is described by Newton's Law of Universal Gravitation,
which states that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically, this can be represented as F
= G * (m1 * m2) / r^2, where F is the gravitational force,
G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers.
As the distance (r) increases, the denominator (r^2) becomes larger, causing the overall force of gravity (F) to decrease.
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a certain wire has resistance r. another wire, of the same material, has half the length and half the diameter of the first wire. the resistance of the second wire is:
The cross-sectional area of the second wire is 1/16th that of the first wire. The resistance of the second wire is four times that of the first wire.
The resistance of a wire is given by the formula:
R = ρ × L / A
where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
Since both wires are made of the same material, their resistivities are the same. Let's denote the length and diameter of the first wire by L1 and D1, respectively, and the length and diameter of the second wire by L2 and D2, respectively. We are given that:
L2 = L1 / 2
D2 = D1 / 2
The cross-sectional area of a wire is given by the formula:
A = π × (D/2)^2
Substituting the given values, we have:
A1 = π × (D1/2)^2
A2 = π × (D2/2)^2
= π × (D1/4)^2
= (1/16) × A1
Therefore, the cross-sectional area of the second wire is 1/16th that of the first wire.
Now we can use the formula for resistance to find the resistance of the second wire:
R2 = ρ × L2 / A2
= ρ × (L1/2) / (1/16 × A1)
= (ρ × L1 × 16) / (2 × A1)
Since the first wire has a resistance of R1, we can substitute its value:
R2 = (ρ × L1 × 16) / (2 × A1)
= (ρ × L1 × 16) / (2 × π × (D1/2)^2)
= 4 × R1
Therefore, the resistance of the second wire is four times that of the first wire.
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the velocity vector of a particle moving in the xy-plane has components given by and . at time , the position of the particle is . what is the y-coordinate of the position vector at time ?
The y-coordinate of the position vector at time t is (5/2)t² - 8t - 1/2.
It is given the velocity vector components as:
v(t) = (4t - 6) i + (5t - 8) j
To find the position vector at time t, we need to integrate the velocity vector with respect to time. integrate each component separately:
x(t) = ∫ (4t - 6) dt = 2t² - 6t + C1
y(t) = ∫ (5t - 8) dt = (5/2)t² - 8t + C2
where C1 and C2 are constants of integration. We can determine these constants by using the initial position of the particle given as:
r(0) = (2, -1)
At time t=0, we have:
x(0) = 2, y(0) = -1
Substituting these values in the expressions for x(t) and y(t), may get:
C1 = 2, C2 = -1/2
So, the position vector at time t is:
r(t) = (2t² - 6t + 2) i + ((5/2)t² - 8t - 1/2) j
To find the y-coordinate of the position vector at time t, may simply need to substitute t into the expression for y(t):
y(t) = (5/2)t² - 8t - 1/2
Therefore, the y-coordinate of the position vector at time t would be (5/2)t² - 8t - 1/2.
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given an average particle diameter of 70mm for the bed of the geography river, what would be the critical tractive force necessary to dislodge that particle?
With the river bed. Assuming the contact area is proportional to the square of the particle's diameter:
F = τc × A
A ≈ D²
To calculate the critical tractive force necessary to dislodge a particle with an average Diameter of 70mm in a river bed, you can follow these steps:
1. Determine the particle's size (D) and convert it to meters: D = 70mm = 0.07m.
2. Calculate the submerged weight (Ws) of the particle. You'll need to know the specific gravity (G) of the particle material and the density of water (ρw). Assuming a typical specific gravity of 2.65 for sediment particles and water density of 1000 kg/m³:
Ws = (G - 1) × ρw × V × g
where V = (4/3)π(D/2)³ is the particle volume and g = 9.81 m/s² is the gravitational acceleration.
3. Calculate the critical shear stress (τc) using the Shields equation:
τc = ρw × g × R × (D/2)
where R is the submerged specific gravity, R = G - 1.
4. Determine the critical tractive force (F) by multiplying the critical shear stress by the particle's contact area (A) with the river bed. Assuming the contact area is proportional to the square of the particle's diameter:
F = τc × A
A ≈ D²
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What is the approximate time difference between the first P-wave and the first S-wave recorded at a seismic station located 8000 kilometers from an earthquake’s epicenter?
*
5 points
8 minutes 40 seconds
9 minutes 20 seconds
11 minutes 20 seconds
20 minutes 40 seconds
The approximate time difference between the first P-wave and the first S-wave recorded at a seismic station located 8000 kilometers from the earthquake’s epicenter would be 11 minutes 20 seconds.
Given the distance from earthquake’s epicenter (d) = 8000km
The approximate time difference between a P-wave and an S-wave can be calculated using the following formula:
From the diagram given we can see that the speed of P-wave (v1)= 8km/s
The speed of S-wave (v2) = 4.75km/s
We know that the time is calculated as distance per speed.
Then time taken for P-wave (t1) = d/v1
Time taken for S-wave (t2) = d/v2
Time Difference (t) = t1 - t2
Then, [tex]t = d/v1 - d/v2 = d((v2 - v1)/v1*v2)[/tex]
[tex]t = 8000 * (8 - 4.75/8 * 4.75)[/tex]
t = 680.4 seconds
So, for an earthquake epicenter located 8000 km away, the time difference would be:
Time Difference = 11 minutes 20 seconds
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flywheel in form of a solid cylinder of mass 60.00 kg and radius 1.80 m is rotated to an angular velocity 26.0 rad/s. what is the energy stored in the flywheel?
The energy stored in the flywheel is approximately 32,949.6 Joules.
The energy stored in the flywheel can be calculated using the formula for rotational kinetic energy: KE = 0.5 * I * ω², where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
For a solid cylinder, the moment of inertia (I) is given by the formula: I = 0.5 * M * R², where M is the mass and R is the radius.
Substituting the given values: I = 0.5 * 60 kg * (1.8 m)² = 97.2 kg m².
Now, we can find the kinetic energy: KE = 0.5 * 97.2 kg m² * (26.0 rad/s)² ≈ 32949.6 J.
So, the energy stored in the flywheel is approximately 32,949.6 Joules.
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the north pole of a magnet is at a set distance from a copper loop which is rotating as shown below. if you are looking at the loop from above the magnet, will you say the induced current is circulating clockwise, counterclockwise, or is equal to 0?
By Lenz's law, the induced current in the copper loop will flow in a clockwise direction when viewed from above the magnet.
Lenz's Law, which stipulates that the direction of the induced current is such that it opposes the change in magnetic flux that caused it, may be used to determine this. The magnetic flux through the copper loop grows as the magnet's north pole gets closer, causing a current to be induced and a magnetic field to be created that opposes the magnetic field that is getting closer. This necessitates that, when viewed from above the magnet, the induced current flow in a clockwise direction.
The magnetic flux through the copper loop also reduces as the magnet's north pole moves away from it, causing a current to flow through the loop and create an opposing magnetic field to the one that is leaving. To counteract the shift in magnetic flux, the induced current must continue to flow in a clockwise direction.
Hence, when viewed from above the magnet, the induced current in the copper loop will move in a clockwise direction.
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Your question is incomplete. The complete question is:
The north pole of a magnet is at a set distance from a copper loop which is rotating as shown below. if you are looking at the loop from above the magnet, will you say the induced current is circulating clockwise, counterclockwise, or is equal to 0? Refer to the image attached to solve the question.
a fire hose exerts a force on the person holding it due to the water accelerating as it goes from the thicker hose out through the narrow nozzle. how much force is required to hold a 7.0-cm-diameter hose delivering through a 0.75-cm-diameter nozzle?
When the water goes from the thicker hose out through the narrow nozzle, a fire hose exerts a force on the person holding it. To hold a 7.0-cm-diameter hose delivering through a 0.75-cm-diameter nozzle, the force required is 34.8 N.
How much force is required to hold a 7.0-cm-diameter hose delivering through a 0.75-cm-diameter nozzle, As per Bernoulli's equation, the pressure P of a fluid (liquid or gas) at any point along its path is equal to the sum of its static pressure (p0),
the kinetic energy per unit volume of the fluid (0.5ρv2),
and its potential energy per unit volume (ρgh),
where ρ is the density of the fluid, v is the velocity of the fluid, g is the acceleration due to gravity, and h is the height relative to a reference point of the fluid point in question. 0.5ρv1^2 + P1 + ρgh1 = 0.5ρv2^2 + P2 + ρgh2
The Bernoulli principle can be simplified to: F1/A1 = F2/A2
Where: F1 and F2 are the forces exerted by the fluid on the hose A1 and A2 are the areas of the hose and nozzle, respectively.
Substituting the values:F1 = (F2A1)/A2 = (A2/A1)ρv2^2A1 = (7/2)2π = 38.48 cm2A2 = (0.75/2)2π = 0.44 cm2v1 = v2 (since the water is incompressible)ρ = 1000 kg/m3Thus:F2 = 0.5ρv2^2A2F1 = 0.5ρv2^2A2 (A1/A2) = (0.5 × 1000 × v2^2 × 0.44) × (38.48/0.44)F1 = 34.8 N
Therefore, the force required to hold a 7.0-cm-diameter hose delivering through a 0.75-cm-diameter nozzle is 34.8 N.
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naphthalene is diffusing from a spherical particle via forced convection into its surroundings. what is the first step in determining its mass flux?
In finding the mass flux of naphthalene after diffusing with a spherical particle, we need to calculate the mass transfer coefficient that can be measured to find out how much better the mass spreads to the surrounding.
The formula for mass transfer is
[tex]Kg/m^{2s} = ( Sh *D)/r[/tex]
Where,
Sh = Sherwood number, D = diffusion coefficient
After calculating the mass transfer using the given formula it is easy to calculate the mass flux using Fick's Law
Fick's Law can be used in forming a formula that can help in finding the mass flux. Therefore,
The formula for Fick's Law is
[tex]J = -Kg/m^{2s} * (C1 - C2)[/tex]
Where
J = mass flux
C1 = presence of naphthalene on the surface of the particle
C2 = presence of naphthalene in the bulk fluid
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a 180-lb (i.e.,5.59 slug) player can accelerate from rest to 15 ft/s in 2 second. if her center of mass accelerates only horizontally, what is the magnitude of the minimum static friction needed so she won't slip? (hint: find her acceleration first)
If her center of mass accelerates only horizontally, the magnitude of the minimum static friction needed so she won't slip is 52 lb.
To find the speed increase of the player, we can utilize the equation a = Δv/Δt, where Δv = 15 ft/s (last speed) and Δt = 2 s (time taken to arrive at that speed). In this way, a = 7.5 ft/s².
To work out the base static erosion expected to forestall slipping, we want to consider the powers following up on the player. The player's weight acts descending (W = mg = 5.59 slug x 32.2 ft/s² = 180 lb), and the power of static erosion acts on a level plane the other way to the course of movement. The greatest power of static grinding is equivalent to the coefficient of static contact (μs) increased by the ordinary power (N = W).
Since the player isn't slipping, the power of static erosion should be equivalent to or more noteworthy than the power expected to evenly speed up her. Accordingly, we can set the power of static contact equivalent to ma*ma (mass x speed increase), which gives:
μsN = ma*ma
μs(W) = ma*ma
μs(180 lb) = (5.59 slug)(7.5 ft/s²)
μs = 0.29
In this manner, the extent of the base static grating required so she won't slip is 0.29 times the heaviness of the player, or around 52 lb.
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Which is a property of every heterogeneous mixture?
a.the mixture is made up of at least two different states.
b.the mixture is made up of something dissolved in a liquid.
c.the composition of the mixture is the same throughout.
d.the characteristics of the mixture change within a sample.
hana fills a cup with sandy ocean water. she pours the mixture through a filter. what does she collect that passes through the filter?
a.a sample of pure water
b.a solution of salt in water
c.a suspension of sand in water
d.a colloid of salt in water
which describe colloids? check all that apply.
1.heterogeneous mixtures
2.homogeneous mixtures
3.may have a uniform appearance
4.are made up of at least two substances
5.will settle out over time
when mixed, which states of matter form only a homogeneous mixture?
a.two liquids
b.two gases
c.a solid and
Answer: C for all of them
Explanation:
Because I'm smart
Jk
So basically its beacuse all of them have the same mixture since science can be interrelated and interchangebale in terms of formulas ok bye now
Thanks
Hope this helped you
Or not
Sorry if it didn't ig
a 2.80 kg grinding wheel is in the form of a solid cylinder of radius 0.100 m .what constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s ?
A constant torque of 0.703 N·m will bring the grinding wheel from rest to an angular speed of 1200 rev/min in 2.5 s.
First, we need to convert the final angular speed to radians per second:
ω = (1200 rev/min) x (2π rad/rev) x (1/60 min/s) = 125.66 rad/s
The moment of inertia of the grinding wheel can be calculated using the formula for a solid cylinder:
I = (1/2)mr² = (1/2)(2.80 kg)(0.100 m)² = 0.014 J·s²
The angular acceleration can be found using the formula:
α = ω/t = (125.66 rad/s) / (2.5 s) = 50.264 rad/s²
The torque required to produce this angular acceleration can be found using the formula:
τ = Iα = (0.014 J·s²)(50.264 rad/s²) = 0.703 N·m
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