"Two conducting spheres have radii of r1 and r2, with r1 greater than r2. If they are far apart the capacitance is proportional to: 1/r1 - 1/r2
The capacitance of two conducting spheres with radii r1 and r2, where r1 is greater than r2, and they are far apart is proportional to the difference in inverse radii. This can be written as:
C = k (A / d)
where C is the capacitance, k is the proportionality constant, A is the surface area, and d is the distance between the two spheres.
The surface area of a sphere is proportional to r^2,
so:C ∝ A = k (r1^2 + r2^2)
The inverse of capacitance is proportional to the difference in inverse radii,
so:1/C ∝ 1/(r1 - r2)
1/C = k' (1/r1 - 1/r2)
where k' is another proportionality constant, which combines with k to give the final constant of proportionality.
Therefore, the capacitance of two conducting spheres with radii r1 and r2, where r1 is greater than r2, and they are far apart is proportional to: 1/r1 - 1/r2.
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How will Newton's 1st and 2nd laws apply to a rocket hitting an asteroid?
Answer:
Newton’s first law states that an object will remain at rest or in motion at a constant velocity unless acted on by a net external force. Newton’s second law states that the force acting on an object is equal to its mass times its acceleration. When a rocket hits an asteroid, the asteroid will remain still until acted upon by the force of the rocket. The force of impact will depend on the mass and acceleration of the rocket.
Explanation:
Newton’s first law, also known as the law of inertia, states that an object at rest remains at rest, or if in motion, remains at a constant velocity unless acted on by a net external force. This means that a rocket will remain still until a force is applied to move it. Once it’s in motion, it won’t stop until a force is applied.
Newton’s second law states that the force acting on an object is equal to its mass times its acceleration (F=ma). This means that the more mass an object has, the more force is needed to move it. During a rocket launch, the engines burn the propellant to fuel the rocket, using it all up until it is gone. Thus, the rocket’s mass becomes smaller as the rocket ascends. So, in keeping with Newton’s second law of motion, the rocket’s acceleration increases as its mass decreases.
When a rocket hits an asteroid, both Newton’s first and second laws come into play. The asteroid will remain still until acted upon by an external force (the rocket). The force of impact will depend on the mass and acceleration of the rocket.
what would have to be the mass of this asteroid for the day to become 22.0% longer than it presently is as a result of the collision? assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.
The required mass of the asteroid in terms of Earth's mass will be 0.088 M.
It is given that the asteroid becomes 22% longer than it is as a result of collision.
From the information given, the length of the day is 2π rad/ω. According to the angular momentum theory,
where,
ω is the Earth's angular rotation rate
Applying conservation of angular momentum to the Earth and asteroid results in,
2/5 M R² ω₁ = (mR² + 2/5 MR²) ω₂
m = 2/5 M(ω₁ - ω₂)/ω₂
T₂ = 1.22T
Above equation can be written as,
1/ω₂ = 1.22/ω₁
ω₁ = 1.22 ω₂
ω₁ - ω₂/ω₂ = 0.22
m = 2/5 (0.22) M = 0.088 M
where,
m is the mass of the asteroid
M is the mass of the earth
Thus, the mass of the asteroid in terms of earth's mass is calculated.
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Two thin circular dics of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length = √24 a through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is co. the angular momentum of the entire assembly about the point 'O' is L (see the figure). Which of the following statement (s) is (are) true?
Answer:
I will say the first one
Explanation:
From the figure, we can see that the assembly consists of two circular discs of different masses and radii, which are rigidly fixed to a massless, rigid rod of length √24 a through their centers. The assembly is set rolling without slipping on a flat surface, and the angular speed about the axis of the rod is co. The angular momentum of the entire assembly about the point 'O' is L.
We can use the conservation of angular momentum to answer the question. Since there is no external torque acting on the system, the angular momentum of the system about point 'O' is conserved. Therefore, the initial angular momentum of the system must be equal to the final angular momentum of the system.
The initial angular momentum of the system can be calculated as follows:
Li = I1 * w1 + I2 * w2
Where:
I1 = moment of inertia of the smaller disc about its center = (1/2) * m * a^2
w1 = angular speed of the smaller disc about its center = co
I2 = moment of inertia of the larger disc about its center = (1/2) * 4m * (2a)^2 = 8ma^2
w2 = angular speed of the larger disc about its center = 0 (since the larger disc is not rotating about its center)
Therefore, the initial angular momentum of the system is:
Li = (1/2) * m * a^2 * co + 8ma^2 * 0
Li = (1/2) * m * a^2 * co
The final angular momentum of the system can be calculated as follows:
Lf = I * wf
Where:
I = moment of inertia of the entire assembly about point 'O' = (1/2) * m * a^2 + (4/3) * m * (2a)^2 = (22/3) * ma^2
wf = final angular speed of the entire assembly about point 'O'
Therefore, the final angular momentum of the system is:
Lf = (22/3) * ma^2 * wf
Since the initial angular momentum must be equal to the final angular momentum, we can equate Li and Lf and solve for wf:
(1/2) * m * a^2 * co = (22/3) * ma^2 * wf
wf = (3/44) * co
Therefore, the final angular speed of the entire assembly about point 'O' is (3/44) times the initial angular speed about the axis of the rod.
From the given options, we can see that statement (i) is true, which states that the final angular speed of the entire assembly is less than the initial angular speed about the axis of the rod. Statement (ii) is false, since the final kinetic energy of the entire assembly is less than the initial kinetic energy about the axis of the rod, due to the work done against friction. Therefore, the correct answer is (i) only.
why is when you jump inside of a train you stay in one place but if you jump on the roof you move further?
When you jump inside of a train, you stay in one place because the train is on rails that guide it along a specific track.
The train is powered by wheels and the rails keep it from leaving the track and moving in any other direction. The wheels also keep you in one place as the train moves forward.
However, if you jump on the roof of the train, you will move further because you are no longer affected by the rails and wheels of the train. You can move in any direction and at any speed as long as you are not affected by the wind, other objects, and the force of gravity. This is why you can move further if you jump on the roof of a train.
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A 2.2 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 200g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick.
a) What is the turntable's angular velocity, in
rpm , just after this event?
Immediately following the blocks' impact, the turntable's angular velocity was around 90.2 rpm.
How quickly does the turntable spin up following this incident, measured in revolutions per minute?We can start out by applying the conservation of angular momentum. The turntable is rotating at an angle of: before the blocks start to fall.
One revolution per minute equals one revolution per hundred rpm, which is equal to 10.47 rad/s.
The equation for the moment of inertia of a solid disc can be used to calculate the moment of inertia of the turntable:
[tex]I = (1/2) m r^2[/tex]
where r is the turntable's radius and m is its mass. By entering the specified values, we obtain:
I is equal to (1/2) (2.2 kg) (0.1 m)2 = 0.011 kg m2.
The blocks stay to the turntable as they strike it, increasing the system's moment of inertia. By combining the inertial moments of the blocks and the turntable, we can determine the new moment of inertia:
[tex]I' = I + 2m(a/2)^2[/tex]
where m is the combined mass of the two blocks (0.2 kg each), an is the turntable's diameter (0.2 m), and the factor 2 takes into consideration the two blocks. When we enter the values, we obtain:
I' = 0.011 kg m2 plus 2 (0.2 kg)(0.1 m)2 = 0.012 kg m2.
where r is the turntable's radius and m is its mass. By entering the specified values, we obtain:
According to the conservation of angular momentum principle, the system's angular momentum is preserved both before and after the blocks strike the turntable. This can be said as follows:
Iω1 = I'ω'
where'is the system's angular velocity immediately following the blocks' impact with the turntable. When we solve for ', we get:
I' = (I1)/I' = (0.011 kg m2)(10.47 rad/s)/(0.012 kg m2)(9.47 rad/s)
This is converted to rpm and we get:
' = 9.47 rad/s times 60 seconds per minute divided by 2 radians per rotation equals 90.2 rpm.
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describe the difference, if any, in the net work done on each sample of gas as it is taken through the cycles shown above. explain how the location of the states on the graphs and the direction of the processes in each cycle can be used to arrive at your answer.
The difference in the net work done on each sample of gas, analyze the location of the states on the graphs, the direction of the processes in each cycle, and the area enclosed by the cycles.
This information will help you compare and arrive at your answer.
The difference in the net work done on each sample of gas in the cycles can be found by examining the location of the states on the graphs and the direction of the processes in each cycle.
Step 1: Identify the cycles on the graphs.
First, recognize the different cycles shown in the graphs. Typically, there are isobaric (constant pressure), isochoric (constant volume), and isothermal (constant temperature) processes involved in these cycles.
Step 2: Analyze the direction of the processes.
Next, look at the direction of the processes in each cycle. Clockwise cycles generally represent positive work done on the gas, while counterclockwise cycles represent negative work done on the gas (or work done by the gas).
Step 3: Calculate the area enclosed by each cycle.
The net work done on the gas in a cycle is equal to the area enclosed by the cycle on the graph. A larger area enclosed would mean more work done, while a smaller area means less work done.
Step 4: Compare the net work done in each cycle.
Now that you have analyzed the area enclosed by each cycle and the direction of the processes, you can compare the net work done on the gas in each cycle. If the cycles have similar areas and the same direction, the net work done in each cycle would be similar. If the areas are different or the cycles have opposite directions, the net work done would be different.
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the magnetic field at the center of a 0.700-cm-diameter loop is 3.00 mt . part a what is the current in the loop? express your answer with the appropriate units.
The magnetic field at the center of a current-carrying loop is given by the formula: B = (μ₀/4π) * (2I/ r)
where B is the magnetic field at the center of the loop, I is the current in the loop, r is the radius of the loop, and μ₀ is the permeability of free space. In this case, we are given the magnetic field B = 3.00 mT = 3.00 × 10^(-3) T and the radius r = 0.700 cm = 0.00700 m.Substituting these values into the above formula, we can solve for the current I: I = (B * r * 4π)/ (2 * μ₀)
I = (3.00 × 10^(-3) T * 0.00700 m * 4π)/ (2 * 4π × 10^(-7) T·m/A)
I = 0.0133 A
Therefore, the current in the loop is 0.0133 A (amperes).
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tarik winds a small paper tube uniformly with 161 turns of thin wire to form a solenoid. the tube's diameter is 7.85 mm and its length is 2.49 cm. what is the inductance, in microhenrys, of tarik's solenoid?
Tarik winds a small paper tube uniformly with 161 turns of thin wire to form a solenoid. the tube's diameter is 7.85 mm and its length is 2.49 cm. The inductance of Tarik's solenoid is approximately 8.858 microhenrys.
To calculate the inductance of Tarik's solenoid, we will use the formula for the inductance of a solenoid:
L = (μ₀ * N² * A) / l
where:
L is the inductance (in henrys),
μ₀ is the permeability of free space (4π × 10⁻⁷ H/m),
N is the number of turns,
A is the cross-sectional area of the solenoid (in square meters),
l is the length of the solenoid (in meters).
1. Convert the given measurements to meters:
Diameter = 7.85 mm = 0.00785 m
Length = 2.49 cm = 0.0249 m
2. Calculate the cross-sectional area (A) using the formula for the area of a circle:
A = π * (d/2)² = π * (0.00785/2)² = 4.835 × 10⁻⁵ m²
3. Plug the given values into the inductance formula:
L = (4π × 10⁻⁷ * 161² * 4.835 × 10⁻⁵) / 0.0249 = 8.858 × 10⁻⁶ H
4. Convert the inductance to microhenrys (1 H = 1,000,000 μH):
L = 8.858 × 10⁻⁶ * 1,000,000 = 8.858 μH
So, the inductance of Tarik's solenoid is approximately 8.858 microhenrys.
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how do spectra show the difference between a type i supernova and a type ii supernova? why does this difference arise? type i supernovae occur when a star composed of
Type I supernovae occur when a star composed mainly of carbon and oxygen accretes matter from a companion star or merges with another white dwarf whereas Type II supernovae are the result of massive stars, containing hydrogen, that has exhausted their nuclear fuel.
Spectra show the difference between a Type I supernova and a Type II supernova by analyzing the elements present in each explosion.
Step 1: Observe the spectra of the supernovae. The spectra display the wavelengths of light emitted by the elements in the explosion.
Step 2: Identify the presence or absence of hydrogen. Type II supernovae have hydrogen lines in their spectra, whereas Type I supernovae do not.
Step 3: Analyze the elements present in Type I supernovae. Type I supernovae are further divided into subcategories based on their spectral lines. For example, Type Ia supernovae show strong silicon lines.
The difference arises due to the nature of the progenitor stars and the mechanisms causing the explosion. When the mass reaches the critical limit, the star undergoes a thermonuclear explosion, resulting in a Type I supernova without hydrogen.
On the other hand, Type II The core collapses under its own gravity, causing a violent explosion that ejects the outer layers, including hydrogen, thus showing hydrogen lines in their spectra.
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A car has a momentum of 6,410 kg m/s and is traveling at 7 m/s. What is the mass of the car?
Explanation:
Momentum = m* v
6410 kg m/s = m * 7 m/s
6410 / 7 = m = 915.7 kg
why does the color of something depend on both the nature of the object itself and the way that we perceive it
The color of an object is determined by the wavelengths of light that it absorbs and reflects.
The chemical composition and physical structure of an object influence how it interacts with light. For example, a red apple appears red because it reflects red wavelengths of light and absorbs other colors. However, the way we perceive color is also influenced by our visual system, which includes our eyes and brain.
Our eyes have photoreceptor cells that detect different wavelengths of light and send signals to the brain, which interprets these signals as colors. Additionally, factors such as lighting conditions and the colors of surrounding objects can affect how we perceive the color of an object. Therefore, both the nature of the object and our visual system contribute to how we perceive its color.
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on motorized solar panel arrays, the panels are moved in order to? select one: a. allow room for maintenance b. track the movement of the sun c. shed snow and ice d. adjust to prevailing wind conditions
On motorized solar panel arrays, the panels are moved in order to track the movement of the sun. The correct answer is option b.
A motorized solar panel array is a device that can adjust the solar panel's orientation to the sun to maximize electricity production. As the sun moves across the sky, photovoltaic solar panels must adjust their orientation to remain perpendicular to the sun's rays.
By tilting or turning solar panels, solar panels can increase energy production. Motorized solar panel arrays enable solar panels to be tracked using an algorithm that considers the time of day, year, and latitude.
Solar panel tracking systems keep solar panels facing the sun to increase energy generation. The most basic tracking systems involve manually adjusting the solar panels' orientation. Motorized solar panel arrays, which automatically adjust the panels' angles based on the time of day, year, and location, are more sophisticated.
Dual-axis tracking systems can adjust the panels both vertically and horizontally. The result is an increase in the amount of energy generated by the solar panels. Therefore option b is correct.
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how much time elapses between the instant when the ball was translating without rotating, and when it rolls without slipping?
Time elapses between the instant when the ball was translating without rotating, and when it rolls without slipping is t = (2/5)MR²μs g/F
Assuming that the ball starts from rest and rolls without slipping at time t = 0, find the time it takes for the ball to start rolling without slipping by using the equations of rotational and translational motion.
Let R be the radius of the ball and I be the moment of inertia of the ball about its center of mass. Since the ball is initially translating without rotating, its angular velocity is zero and its linear velocity is given by:
v = rω
where r is the distance from the center of mass to the point of contact with the ground, which is equal to R in this case.
The acceleration of the ball is given by:
a = αR
where α is the angular acceleration of the ball. Since the ball is initially not rotating, α is zero. However, as the ball starts to roll without slipping, a frictional force acts on it, causing it to rotate. The torque due to this force is given by:
τ = Fr = Iα
where F is the magnitude of the frictional force.
At the instant when the ball starts to roll without slipping, the linear velocity and angular velocity are related by:
v = Rω
And the acceleration and angular acceleration are related by:
a = Rα
Since the ball is rolling without slipping, the linear velocity and angular velocity are related by:
v = Rω
And the acceleration and angular acceleration are related by:
a = Rα
Can use these equations to find the time it takes for the ball to start rolling without slipping. At this instant, the frictional force has reached its maximum value and is equal to the force of static friction, given by:
F = μsmg
where μs is the coefficient of static friction between the ball and the ground, and mg is the weight of the ball.
Thus, can write:
τ = Fr = Iα = μsmgR
Substituting Rω for v and Rα for a, can be get:
μsmgR = Iα/R = I(Rω)/R²
Solving for ω, we get:
ω = μs g R/I
The time it takes for the ball to start rolling without slipping is the time it takes for the angular velocity to reach this value. Using the equation:
ω = αt
t = ω/α = Iμs g R/F
Substituting the values for the moment of inertia, radius, coefficient of static friction, and weight of the ball, get:
t = (2/5)MR²μs g/F
where M is the mass of the ball.
Thus, the time it takes for the ball to start rolling without slipping depends on the mass of the ball, the radius of the ball, the coefficient of static friction between the ball and the ground, and the magnitude of the weight of the ball.
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the acceleration due to gravity at the surface of the moon is 1.63 m/s2. what is the weight of an astronaut standing on the moon whose weight on earth is 206 lb?
The acceleration due to gravity at the surface of the moon is 1.63 m/s2. The weight of the astronaut standing on the moon whose weight on earth is 206 lb would be approximately 153 Newtons.
Weight on the Moon = (Weight on Earth) x (Acceleration due to gravity on the Moon) / (Acceleration due to gravity on Earth)
we can convert the weight on Earth from pounds to Newtons (since weight is a force and is measured in Newtons):
Weight on Earth = 206 lb x 4.45 N/lb
Weight on Earth = 917.7 N
we can substitute the given values into the formula:
Weight on the Moon = (917.7 N) x (1.63 m/s²) / (9.81 m/s²)
Weight on the Moon ≈ 153 N
Therefore, the weight of the astronaut standing on the moon would be approximately 153 Newtons.
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Look at the graph. What is the slope of the line?
Answer:D
Explanation: you’re going in a downward direction. Consider rise/run.
Like most spacecraft returning from orbit, the Apollo command module entered the atmosphere at 7.8 km/s. In front of the capsule was a shock front, the leading edge of the shock front we call a bow shock. Let's consider the conditions as it passes an altitude of 50 miles, at about 17,500 miles per hour. What's the altitude, in meters, the speed in km/s ?
The provided student question, we can say that at an altitude of 50 miles, the Apollo command module would be traveling at a speed of 7.8 km/s, or 7800 m/s. The altitude at this speed and altitude would be approximately 80,470 meters.
When answering questions on Brainly, it is important to always be factually accurate, professional, and friendly. Answers should be concise and not provide extraneous amounts of detail. Typos and irrelevant parts of the question should be ignored. Additionally, using the terms provided in the question can help to ensure that the answer is relevant and helpful to the student.
In response to the provided student question, at an altitude of 50 miles, the Apollo command module would be traveling at a speed of approximately 7.8 km/s. To convert this speed to meters per second, we can use the conversion factor of 1 km/s = 1000 m/s. Therefore, the speed of the module in meters per second would be:
7.8 km/s * 1000 m/s = 7800 m/s
The altitude of the module at this speed and altitude would be approximately 80.47 kilometers, or 80,470 meters. This calculation can be done using the formula for atmospheric density, which takes into account factors such as temperature, pressure, and composition of the atmosphere. At an altitude of 50 miles,
the density of the atmosphere is relatively low, which allows the module to maintain a high speed without burning up due to friction with the air.
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how can a stationary metal sphere have kinetic energy, the energy of motion?(1 point) responses the metal is made of atoms, which are vibrating in place.
Kinetic energy refers to the gain of energy in response to the movement of an object. It depends on the mass and velocity at which it is moving from one point to another in a specific period. The SI unit is Joule(J). The formula is K.E = [tex]\frac{1}{2}[/tex] m × v² .
Furthermore, the generation of Kinetic energy takes place in the metal sphere at an atomic level due to the constant vibration that takes place between them in a specified time. This phenomenon occurs because the Kinetic energy not only focuses on the metal sphere at a base level but also at subatomic level.
This generation of Kinetic energy is minute between the atoms but it isn't zero.
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how much work must you do to push a 12 kg k g block of steel across a steel table at a steady speed of 1.3 m/s m / s for 6.1 s s ? the coefficient of kinetic friction for steel on steel is 0.60.
To determine the work necessary to drive a 12 kilogram block of steel over a steel table at a constant speed of 1.3 m/s for 6.1 seconds, we must take into account the force required to overcome friction.
Steel against steel has a coefficient of kinetic friction of 0.60.Then, we must compute the force of friction opposing the block's motion. F friction = coefficient of friction x F normal, where F normal is the normal force applied by the table to the block. We know that the net force applied on the block is zero since it is travelling at a constant pace. As a result, the force of friction must be equal to the force we are exerting.to move the block ahead. Applying the aforementioned equation, we can calculate the friction force to be 70.56 N. As a result, the effort required to push the block for 6.1 seconds is equal to the friction force multiplied by the distance the block moves during that time, which is given by distance = speed x time = 1.3 m/s x 6.1 seconds = 7.93 m. Hence, W = force x distance = 70.56 N x 7.93 m = 560.3 J is the labor necessary to push the block (Joules).
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A snow monster drags his injured prey across the snow. Assume
friction is zero. If the prey has a mass of 85 kg. If the monster applies
a constant force of 261 N at an angle of 60 degrees to drag the prey
32 m. How much work did the monster do?
The amount work done by the monster is 4176 J.
What is work done?Work is said to be done when a force moves an object through a distance.
To calculate the work done by the monster, we use the formula below
Formula:
W = Fdcos∅........................ Equation 1Where:
W = Work done by the monsterF = Force applied by the monsterd = Distance∅ = Angle to the horizontalFrom the question,
Given:
F = 261 Nd = 32 m∅ = 60°Susbtitute these values into equation 1
W = 261×32×cos60°W = 4176 JHence, the work done is 4176 J.
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you can move the bar magnet around. when does the light bulb shine brightest with the loop area fixed?
The light bulb shines brightest when the bar magnet is moved quickly in and out of the loop area.
The light bulb will shine brightest when the bar magnet is moved quickly in and out of the loop area. This is because the faster the magnet moves, the greater the change in magnetic field and the greater the induced emf and current in the loop.
At the same time, the resistance of the loop remains constant, so the power dissipated by the bulb, which is proportional to the square of the current, increases with the speed of the magnet. If the magnet is moved too slowly, the induced emf and current will be too weak to light up the bulb.
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a certain transformer has an input voltage of 127 vac, an output voltage of 26.5 vac and 122 turns in the secondary coil. how many turns are there in the primary coil?
There are approximately 587 turns in the primary coil of this transformer.
To determine the number of turns in the primary coil of a transformer with an input voltage of 127 VAC, an output voltage of 26.5 VAC, and 122 turns in the secondary coil, you can use the transformer equation:
Vp / Vs = Np / Ns
where Np is the number of turns in the primary coil, Vp is the primary voltage, Vs is the secondary voltage, and there are Np turns in the primary coil and Ns turns in the secondary coil. Inputting the numbers provided yields:
127 / 26.5 = Np / 122
Simplifying the equation by multiplying both sides by 122, we get:
Np = (127 / 26.5) x 122
Np = 586.8
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what is the wavelength of a wave that has a speed of 350 Meters/second and a frequency of 140 hertz?
____ meters
Answer:
[tex] \lambda = 2.5 \: m[/tex]Explanation:
To find:-
The wavelength of the wave.Answer:-
We are here given that , the speed of a wave is 350m/s and has a frequency of 140Hz . We are interested in finding out the wavelength of the wavelength of the wave .
As we know that, wavelength, frequency and speed are related to each other as ,
[tex]\longrightarrow\boxed{ v = \lambda \nu} \\[/tex]
where,
[tex]v[/tex] is the speed of the wave.[tex]\lambda[/tex] is the wavelength of the wave.[tex]\nu[/tex] is the frequency of the wave.On substituting the respective values, we have;
[tex]\longrightarrow 350 m/s = \lambda \times 140Hz \\[/tex]
[tex]\longrightarrow \lambda =\dfrac{350}{140} m \\[/tex]
[tex]\longrightarrow \boxed{\boldsymbol \lambda = 2.5 \ m} \\[/tex]
Hence the wavelength of the wave is 2.5 m .
Answer:
2.5meter
Explanation:
edge 2023
a ball is attached to a string that is staked to the center of a frictionless table and the ball circles around the tabletop as shown below. a black ball is attached to a string that is staked at the the center of a table. a horizontal-oval-grey-dashed arrow points counter clockwise above the table. a black ball is attached to a string that is staked at the the center of a table. a horizontal-oval-grey-dashed arrow points counter clockwise above the table. which set of force vectors shows all the horizontal forces on the ball?
A rope with a dark ball connected to it is anchored in the middle of a table. The set of force vectors that shows all the horizontal forces on the ball is zero.
As the ball is moving in a circular path, it experiences a centripetal force, which is directed towards the center of the circle, perpendicular to the velocity of the ball. Therefore, the set of force vectors that shows all the horizontal forces on the ball is zero. No horizontal forces are acting on the ball.
The only force acting on the ball is the tension force provided by the string which acts radially towards the center of the circle. As there is no net force in the horizontal direction, the ball continues to move in a circular path at a constant speed, governed by the radius of the circle and the angular velocity.
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The mass of a train is 120,000 kg. Calculate the unbalaced force on the train at 10s.
The unbalanced force would be: F = (120,000 kg) * (5 m/s^2) = 600,000 N
To calculate the unbalanced force on the train at 10 seconds, we need to know the acceleration of the train and the forces acting upon it.
Using Newton's second law of motion, we know that the unbalanced force on an object is equal to the product of its mass and acceleration.
F = ma
We are given the mass of the train, which is 120,000 kg. To determine the acceleration of the train, we need to know the net force acting on it.
If we assume that the train is moving at a constant velocity, we can conclude that the net force acting on it is zero. This is because the train is not accelerating, so the forces acting on it must be balanced.
However, if the train is accelerating or decelerating, there must be an unbalanced force acting on it. This force could come from a number of sources, such as the engine or the brakes.
If we are given the acceleration of the train, we can calculate the unbalanced force using the equation above. For example, if the train is accelerating at a rate of 5 m/s^2, the unbalanced force would be:
F = (120,000 kg) * (5 m/s^2) = 600,000 N
If we are not given the acceleration of the train, we would need to use additional information to determine it, such as the speed of the train and the distance it travels in a certain amount of time.
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use the measured initial velocity to predict the final velocity with sign (hint: eqs. (7)- (9) might be helpful). compute the % difference between the measured and predicted velocities. the % difference is defined as
To predict the final velocity with sign, use the measured initial velocity and equations (7)-(9). Then compute the percent difference between the measured and predicted velocities.
To predict the final velocity with sign using the measured initial velocity, you can follow these steps:
Step 1: Identify the relevant equations from the hint provided. Equations (7)-(9) might be helpful.
Step 2: Determine the initial velocity and any other necessary information from the problem statement or data provided.
Step 3: Use the appropriate equation from Step 1 to predict the final velocity with sign. This may involve using the acceleration, time, or distance traveled in the calculation.
Step 4: Calculate the percentage difference between the measured and predicted velocities using the formula:
Percentage Difference = (|Measured Velocity - Predicted Velocity| / Measured Velocity) * 100
Make sure to use the absolute value of the difference to ensure a positive percentage.
By following these steps, you will be able to predict the final velocity with sign and compute the percentage difference between the measured and predicted velocities.
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adrienne (50kg) and bo (90 kg) are playing on a 100 kg rigid plank resting on supports. if adrienne stands on the left end, can bo walk all the way to the right end without the plank tipping over? if not, how far can he get past the support on the right?
The distance from the left support is greater than 49/9 meters, then the plank will tip over.
When responding to questions on the Brainly platform as a question-answering bot, it is important to be factually accurate, professional, friendly, and concise. Irrelevant parts of the question or typos should not be ignored.
Additionally, using relevant terms from the question in the answer is important to ensure that the answer is clear and specific.Here is a possible answer to the student question: Adrienne (50 kg) and Bo (90 kg) are playing on a 100 kg rigid plank resting on supports.
If Adrienne stands on the left end, can Bo walk all the way to the right end without the plank tipping over? If not, how far can he get past the support on the right?Assuming that the plank is uniform and mass is distributed uniformly,
the weight of the plank can be taken to be halfway between Adrienne and Bo, which is 70 kg. Using the principle of moments, the torque due to Adrienne is equal to the torque due to Bo when the plank is balanced. Thus,
the product of the weight and distance of Adrienne from the pivot point (left support) is equal to the product of the weight and distance of Bo from the pivot point. Therefore,50 kg × d1 = 90 kg × d2where d1 and d2 are the distances of Adrienne and Bo from the left support, respectively.
The total distance of the plank from left to right is L, which is the sum of the distances from the left support to the right support, and is 2d2. Thus, L = 2d2.
When the plank is about to tip over, the torque due to Bo is equal to the maximum torque that the plank can withstand before tipping over, which is the product of half the weight of the plank and half the length of the plank, that is,70 kg × L/2.
Therefore,90 kg × d2 = 70 kg × L/2.
Substituting L = 2d2,
we obtain90 kg × d2 = 70 kg × 2d2/2= 70 kg × d2Bo can walk all the way to the right end without the plank tipping over if his distance from the left support, d2, is less than or equal to 49/9 meters (approximately 5.44 meters).
This means that he can walk 50/9 meters (approximately 5.56 meters) past the right support before the plank tips over.
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a person's near point is 25 cm, and her eye lens is 2.7 cm away from the retina. what must be the focal length of this lens for an object at the near point of the eye to focus on the retina?
The focal length of the lens must be 62.5 cm for an object at the near point of the eye to focus on the retina.
The near point of the eye is the closest distance at which the eye can focus on an object. Let's call this distance "p". The distance between the lens and the retina is called the "ocular distance" and is denoted by "d". According to the thin lens equation, the focal length "f" of the lens is related to the object distance "p" and the image distance "q" by the equation,
1/f = 1/p + 1/q
Since the object is at the near point of 25 cm, we can set p = 25 cm. The image distance q is the distance between the lens and the retina, which is d = 2.7 cm. Solving for f,
1/f = 1/p + 1/q
1/f = 1/25 + 1/2.7
1/f = 0.04
f = 25 cm/0.04
f = 62.5 cm
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christopher and his mother went on a camping trip. his mother struck a match to light a lantern. which forms of energy did the burning match give off? responses nuclear and solar nuclear and solar solar and thermal solar and thermal thermal and light thermal and light nuclear and light
Thermal and light energy are the two types of energy emitted by a burnt match. On their camping vacation, Christopher's mother struck a match to light a lantern, and the burning match released two types of energy.
heat and light. Once the matchstick began to burn, the chemical energy held in it was transformed into thermal energy, generating heat in the process. The heat then drove the surrounding air molecules to vibrate, generating thermal energy that Christopher and his mother could feel. The burning match emitted light energy in addition to temperature energy. The chemical processes that occurred while the matchstick burnt created light energy in the form of a flame. Christopher and his mother were able to perceive this light energy, which gave the lighting required to light the lantern. the burning match gave off both thermal and light energy, which are common forms of energy released during combustion reactions.
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6. What type of precipitation have ice crystals melt as they fall through a slightly warmer layer of air? Then, they refreeze into small ice pellets as they pass through a cold layer of air closer to the ground.
The type of precipitation described in this scenario is called sleet. It occurs when snowflakes melt into raindrops in a layer of warm air, and then refreeze into ice pellets before reaching the ground in a layer of cold air closer to the surface.
Sleet is often associated with winter storms and can make roads and sidewalks slippery and hazardous.Sleet is the term for the process you are describing. Sleet is a type of precipitation that develops when snowflakes travel through a warm air layer and partially melt before falling through a cold air layer closer to the ground and refreezing as ice pellets. This typically occurs when there is a warm air layer above and a cold air layer close to the surface. When snowflakes fall through warm air, they start to melt, but if the temperature is below freezing close to the ground, they will refreeze into ice pellets before they reach the surface. Sleet differs from freezing rain, which happens when droplets come into contact with a cold surface, like the ground, and then freeze.
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a small rubber wheel is used to drive a large pottery wheel. the two wheels are mounted so that their circular edges touch. the small wheel has a radius of 2.0 cm and accelerates at the rate of , and it is in contact with the pottery wheel (radius 27.0 cm) without slipping. calculate (a) the angular acceleration of the pottery wheel, and (b) the time it takes the pottery wheel to reach its required speed of 65 rpm.
The angular acceleration of the pottery wheel is 400 rad/[tex]s^2[/tex] and it takes 0.922 seconds to reach 65 rpm.
This issue includes the connection among straight and rakish speed increase and speed. The little elastic wheel has a sweep of 2.0 cm and advances at a pace of 8.0 m/[tex]s^2[/tex]. Utilizing the condition a = rα, we can work out that the rakish speed increase of the stoneware wheel is likewise 400 rad/[tex]s^2[/tex].
To make the opportunity it takes the earthenware wheel to arrive at its expected speed of 65 rpm, we first proselyte 65 rpm to radians each second utilizing the condition ω = (rpm)(2π/60). We then utilize the condition v = rω to find the direct speed of the ceramics wheel, which we use in the situation θ = ωt to make the opportunity it takes to pivot one full transformation. The last response is roughly 0.922 seconds.
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