Answer:
Incresing the staircase does not affect the work done; only the vertical distance affects the work done. Remember, the work is done against the weight and thus the relevant displacement is only the vertical displacement.
Explanation:
what has mobed to produce the charge on the plates
Charges need to be transferred from one plate to the other to generate a charge on a capacitor's plates. Electrons are usually moved from one plate to the other by attaching the plates to a power source, such as a battery.
The plate attached to the negative terminal of the battery becomes negatively charged and the plate connected to the positive terminal of the battery becomes positively charged when the power source is connected.
Until the potential difference between the plates equals the potential difference between the battery connections, electrons move from the negative plate to the positive plate. The flow of electrons stops at this moment, and the capacitor is fully charged.
In an electric field that is created between the plates of a capacitor, the charges on the plates are retained. The potential differential between the plates is produced by this electric field and is inversely proportional to the charge on the plates and their separation from one another.
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how the kinetic frictional force on the wooden block depends on the normal force on the wooden block
The kinetic frictional force on a wooden block is directly proportional to the normal force on the block. This means that as the normal force increases, so does the kinetic frictional force, and vice versa.
The kinetic frictional force is the force that opposes the motion of a sliding object. When an object, such as a wooden block, is in contact with a surface, there is a normal force acting on the object that is perpendicular to the surface. The normal force is the force that prevents the object from passing through the surface. The frictional force between the object and the surface is directly proportional to the normal force. This is known as Coulomb's law of friction. The coefficient of friction is a constant that determines the amount of frictional force between the object and the surface. The coefficient of kinetic friction is used to calculate the frictional force when the object is in motion. It is a ratio of the kinetic frictional force and the normal force. Therefore, the kinetic frictional force on a wooden block depends on the normal force on the block and the coefficient of kinetic friction between the block and the surface. As the normal force increases, the frictional force also increases proportionally.
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Place the following regions of the Sun in order of increasing radius
a. corona
b. core
c. radiative zone
d. convective zone
e. chromosphere
f. photosphere
g. a sunspot
The core, radiative zone, convective zone, photosphere, chromosphere, and corona are the sun's regions, listed in order of increasing radius from the centre out.
The order of these regions is core, radiative zone, convective zone, photosphere, chromosphere, and corona, from smallest to greatest. With temperatures ranging from 10-15 million degrees Celsius, the core of the sun, which is its innermost layer, is also its hottest region.Here, nuclear fusion occurs, transforming hydrogen into helium and liberating a massive amount of energy. Temperatures in the radiative zone, which surrounds the core, range from 1-2 million degrees Celsius. Here is where the core's energy is sent outside the body. It is the convective zone that temperatures here vary from 1-2 million degrees Celsius in the layer underneath the photosphere.Convection is the method used to move hot gas to the surface at this location. The visible layer of the sun is called the photosphere, and it is also its brightest region. Here, the temperature varies between 5,000 and 6,000 degrees Celsius.The chromosphere, which sits on top of the photosphere, has a temperature of 20,000 degrees Celsius, making it significantly hotter than the photosphere. The corona, which is the sun's outermost layer and its hottest region, reaches temperatures of 2-3 million degrees Celsius. This layer, which is made up of hot, ionized gas, is where solar winds originate.To know more about chromosphere
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an object is in uniform circular motion. it is traveling at a constant speed. is the net force acting on the object zero? why or why not?
When an object is in uniform circular motion and traveling at a constant speed, the net force acting on the object is not zero.
It is not zero because an object in uniform circular motion is constantly changing direction, even if its speed remains the same. This change in direction requires a force, which is provided by centripetal force. Centripetal force is the force that keeps an object moving in a circular path by pulling it toward the center of the circle.
This force is always directed toward the center of the circle and is perpendicular to the object's velocity. Without this force, the object would continue moving in a straight line.
Thus, the net force acting on an object in uniform circular motion is not zero, but rather is equal to the centripetal force.
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a baseball collides with a baseball glove. which equation is used to calculate the force the glove exerts on the ball during the collision?
A baseball and a glove make contact. What formula is used to determine how much force the glove applies to the ball when it collides is W= f x d.
The ball striking the glove would be the action force, and the glove applying force back to the ball would be the reaction force.
W = f x d
W = work done is the formula for work completed.
Force = F, and Distance = D
The forces are balanced because the baseball is at rest in the pitcher's glove. When the ball is moving during the pitcher's windup and release, the forces are out of balance. Then, as the ball continues to move in the air at a consistent speed, they balance themselves once more. When the ball slows down after hitting the catcher's gloves, the forces become out of balance. Once the ball is absolutely still in the catcher's glove, they balance out again.
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Which is the answer to this question???
The student group that built the strongest electromagnet is (option C) Group 4, because the strength of the electromagnet depends directly on the number of loops, current in the circuit and the presence of a core.
What is electromagnet?An electromagnet is a type of magnet that is created by passing an electric current through a coil or wire. The electric current generates a magnetic field, which can be enhanced by using a magnetic core material, such as iron. Electromagnets can be turned on and off by controlling the flow of electric current through the wire.
They are used in a wide range of applications, from simple devices like doorbells and magnetic locks to more complex technologies like MRI machines and particle accelerators.
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A mass m=15.0 kg is pulled along a horizontal floor, with a coefficient of kinetic friction μk =0.06, for a distance d=8.5 m. then the mass is continued to be pulled up a frictionless incline that makes an angle θ=35.0° with the horizontal. the entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ=35.0° (thus on the incline it is parallel to the surface) and has a tension t=45.0 n.1) what is the work done by tension before the block gets to the incline?2) what is the work done by friction as the block slides on the flat horizontal surface?
(a) the work done by tension before the block gets to the incline: Wt = 199.2 J. (b) the work done by friction as the block slides on the flat horizontal surface: Wk = - 45.5 J.
a)
consider the motion along the horizontal direction
T = tension force in the rope pulling the mass = 33 N
d = displacement of the mass before it gets to the inclined surface = 6.9 m
ϴ = angle between the tension force and displacement = 29
work done by tension force is given as
[tex]w_{t}[/tex] = work done by tension force = T d Cosϴ
[tex]w_{t}[/tex] = (33 x 6.9) Cos29
[tex]w_{t}[/tex] = 199.2 J
b)
for the mass, the force equation in the vertical direction is given as
T Sinϴ + Fn = mg
inserting the values
33 Sin29 + Fn = 15 x 9.8
Fn = 131 N
uk = Coefficient of kinetic friction = 0.05
the kinetic frictional force is given as
[tex]f_{k} = u_{k} f_{n}[/tex]
inserting the values
[tex]f_{k}[/tex] = (0.05) (131)
[tex]f_{k}[/tex] = 6.6 N
Ф = angle between the displacement "d" and frictional force "[tex]f_{k}[/tex]" = 180
Wk = [tex]f_{k}[/tex] d Cos\Ф
work done by the frictional force is given as inserting the values
Wk = (6.6) (6.9) Cos180
Wk = - 45.5 J
c)
v = speed gained by the mass before it gets to the incline
using work-change in kinetic energy theorem
work done by the external force = change in kinetic energy
Wt + Wk = (0.5)m v2
199.2 + (- 45.5) = (0.5) (15) v2
v = 4.5 m/s
d)
a = acceleration of mass parallel to the incline
consider the motion parallel to the inclined surface
parallel to the incline, the force equation for the motion is given as
T - mg Sin29 = ma
33 - (15 x 9.8) Sin29 = 15 a
a = - 2.6 m/s2
D = distance traveled parallel to the incline before coming to rest
vi = initial velocity at the start of incline = v = 4.5 m/s
v = final velocity as it comes to stop = 0 m/s
using the kinematics equation
vf2 = vi2 + 2 a D
02 = 4.52 + 2(- 2.6) D
D = 3.89 m
e)
h = height gained by the mass on an incline
Sin29 = h/D
hence h = D Sin29
h = 3.89 Sin29
h = 1.89 m
Fg = force of gravity in downward direction = mg = 15 x 9.8 = 147 N
= angle between the force of gravity "Fg" in the down direction and displacement "h" in the upward direction = 180
Work done by gravity is given as
Wg = Fg d Cosα
Wg = (147) (1.89) (Cos180)
Wg = - 277.83 J
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the complete question is:
A mass m = 15 kg is pulled along a horizontal floor, with a coefficient of kinetic friction. k = 0.05, for a distance d = 6.9 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle. = 29?½ with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of? = 29?½ (thus on the incline it is parallel to the surface) and has a tension T = 33 1)What is the work done by tension before the block gets to the incline? 2)What is the work done by friction as the block slides on the flat horizontal surface? 3)What is the speed of the block right before it begins to travel up the incline? 4)How far up the incline does the block travel before coming to rest? 5)What is the work done by gravity as it comes to rest?
A solid cylinder with mass M. radius R, and rotational inertia 1/2MR² rolls without slipping down the inclined plane
shown above. The cylinder starts from rest at a height H. The inclined plane makes an angle with the horizontal.
Express all solutions in terms of M, R, H, theta, and g.
a. Determine the translational speed of the cylinder when it reaches the bottom of the inclined plane.
b. Show that the acceleration of the center of mass of the cylinder while it is rolling down the inclined plane is (2/3)g sin theta.
c. Determine the minimum coefficient of friction between the cylinder and the inclined plane that is required for the cylinder to roll without slipping.
The acceleration of the center of mass of the cylinder is (2/3)g sinθ, and the minimum coefficient of friction between the cylinder and the inclined plane is μ_min = tanθ.
What is accelaration?Acceleration is the rate of change of velocity of an object. It measures the rate at which an object's speed and direction changes. It is a vector quantity, meaning it has both magnitude and direction. Acceleration can be positive, negative, or zero. Positive acceleration indicates an object is speeding up, while negative acceleration indicates it is slowing down. Zero acceleration indicates an object is maintaining its current speed and direction.
Determine the translational speed of the cylinder when it reaches the bottom of the inclined plane.
b. Show that the acceleration of the center of mass of the cylinder while it is rolling down the inclined plane is (2/3)g sin theta.
c. Determine the minimum coefficient of friction between the cylinder and the inclined plane that is required for the cylinder to roll without slipping.
Answer:
a. The translational speed of the cylinder when it reaches the bottom of the inclined plane is given by v = (2gh sinθ/3)^1/2.
b. The acceleration of the center of mass of the cylinder is given by a = (2/3)g sinθ. This can be shown as follows: The acceleration due to gravity is g cos θ, so the normal force is Mg cos θ. The rotational inertia of the cylinder is 1/2MR², and its angular acceleration is a/R. By Newton's second law, the resultant force acting on the cylinder is ma = Mg cosθ - 1/2MR² a/R. Therefore, a = (2/3)g sinθ.
c. The minimum coefficient of friction between the cylinder and the inclined plane required for the cylinder to roll without slipping is μ_min = tanθ. This can be shown as follows: The force of friction acting between the cylinder and the inclined plane is given by F_f = μN = μMg cos θ. The force of friction must be equal to the centripetal force acting on the cylinder, which is Mv²/R. Therefore, μ = v²/gR = (2gh sinθ/3)^1/2/gR = tanθ.
In conclusion, we have shown that the translational speed of the cylinder when it reaches the bottom of the inclined plane is given by v = (2gh sinθ/3)^1/2, the acceleration of the center of mass of the cylinder is (2/3)g sinθ, and the minimum coefficient of friction between the cylinder and the inclined plane is μ_min = tanθ.
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what frequency is detected by a stationary train? the velocity of sound is 343 m/s . answer in units of hz.
Sοund with frequency f0=492 Hz is emitted frοm a statiοnary sοurce. A big car mοving at a 2 ms-1 speed tοward the sοund sοurce reflects the sοund.
Hοw fast is that, fοr instance?The rate οf sοmething mοving in οne directiοn is called its velοcity. As an illustratiοn, think οf the velοcity οf a car driving nοrth οn such a highway οr the speed at which a rοcket takes οff. Because the velοcity vectοr is scalar, it always has the same absοlute value magnitude as the mοtiοn's speed.
Is speed always equal tο velοcity?Only when a mοving bοdy mοves inside a single uninterrupted path dο the speed and velοcity measurements match in magnitude. Nοnetheless, if a bοdy dοesn't mοve in a single, straight line.
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Complete Question:
What frequency is detected by a stationary train? The velocity of sound is 343 m/s .
Answer in units of Hz.
3. Calculate the electric force that exists between two objects that are 5. 0 x
10-2 m apart and carry charges of 2. 5 x 10-6 C and −3. 2 × 10¯6 C. Is this force
attractive or repulsive?. (1 point)
-6
The electric force between the two objects with charges of 2.5 × 10⁻⁶ C and -3.2 x 10⁻⁶ C and a distance of 5.0 × 10⁻² m is calculated to be -4.608 N, indicating an attractive force between them.
The electric force between two charged objects is given by Coulomb's Law, which states that:
F = kq₁q₂ / r²
where F is the electric force, q₁ and q₂ are the charges of the two objects, r is the distance between them, and k is Coulomb's constant, which has a value of approximately 9.0 × 10⁹ N·m²/C².
Plugging in the given values, we get:
F = (9.0 × 10⁹ N·m²/C²) * (2.5 × 10⁻⁶ C) * (-3.2 × 10⁻⁶ C) / (5.0 × 10⁻² m)²
F = -4.608 N
The negative sign indicates that the force is attractive, meaning that the two objects will be pulled towards each other.
Therefore, the electric force between the two objects is 4.608 N and it is attractive.
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Comlpete question:
Calculate the electric force that exists between two objects that are 5.0 × 10⁻² m apart and carry charges of 2.5 × 10⁻⁶ C and -3.2 x 10⁻⁶ C. Is this force attractive or repulsive?
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the density at any point is proportional to the square of its distance from the origin. (Assume that the coefficient of proportionality is k. )
To find the moments of inertia Ix, Iy, I0 for the given lamina, we first need to calculate its mass and centroid. The density at any point is proportional to the square of its distance from the origin,
so the mass element dm can be expressed as kr^2dA, where k is the coefficient of proportionality, r is the distance from the origin, and dA is the differential area element. Using polar coordinates, we can express the given region as 0 ≤ r ≤ 6 and 0 ≤ θ ≤ π/2. Integrating dm over this region, we get the total mass of the lamina as: M = ∫∫ kr^2dA = k ∫∫ r^2dA = k ∫θ=0..π/2 ∫r=0..6 r^2r dr d = k ∫θ=0..π/2 [r^4/4]_r=0..6 dθ = (3/5)πk(6^5) To find the centroid of the lamina, we can use the formulae: x_c = (1/M) ∫∫ xdm, y_c = (1/M) ∫∫ ydm Simplifying, we get: x_c = (1/M) k ∫∫ xr^2dA, y_c = (1/M) k ∫∫ yr^2dA.
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a uniform electric field points along the x axis. if a stationary electron is placed in this field, in what direction will it be forced to sstart to move?
The electron will be forced to move in the direction of the electric field, which in this case is the positive x-axis direction.
The electric field exerts a force on the electron that is equal to the product of the charge of the electron and the magnitude of the electric field. The force will act to accelerate the electron in the direction of the electric field.
An electron that is stationary will be forced to move in the direction opposite to that of the electric field if it is placed in a uniform electric field that points along the x-axis. The direction of the electric field determines the direction of the force felt by the charged particle.A uniform electric field is one in which the electric field is the same at all points in space. The strength of the electric field, in this case, is independent of the position of the point in space. The electric field is a vector quantity that is directed along the direction of the force experienced by the charged particle that is placed in the field.
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a flea jumps by exerting a force of straight down on the ground. a breeze blowing on the flea parallel to the ground exerts a force of on the flea while the flea is still in contact with the ground. find the direction and magnitude of the acceleration of the flea if its mass is . do not neglect the gravitational force
The direction of acceleration of the flea is upward and the magnitude of acceleration is given by: a = [tex]\frac{F_{breeze} - mg}{m}[/tex] where [tex]F_{breeze}[/tex] is the force exerted by the breeze on the flea, m is the mass of the flea, g is the acceleration due to gravity, and a is the resulting acceleration of the flea.
When the flea jumps, it exerts a force straight down on the ground, which according to Newton's third law, results in an equal and opposite force exerted by the ground on the flea, causing it to move upward. However, the breeze blowing parallel to the ground exerts a force on the flea in the opposite direction to its motion, which reduces the upward force exerted by the ground, and hence the acceleration of the flea.
To calculate the magnitude of acceleration of the flea, we need to find the net force acting on the flea. This is given by the difference between the force exerted by the breeze and the gravitational force acting on the flea, which is given by mg. The direction of the net force is upward since the force exerted by the breeze is in the opposite direction to the motion of the flea.
The magnitude of acceleration can then be calculated using Newton's second law, a =[tex]{F_{net}/m[/tex], where[tex]F_{net}[/tex] is the net force and m is the mass of the flea.
Therefore, the direction of acceleration of the flea is upward and the magnitude of acceleration is given by: a = [tex]\frac{F_{breeze} - mg}{m}[/tex] where the force exerted by the breeze, F_breeze, and the mass of the flea, m, are given in the problem statement. The acceleration due to gravity, g, is approximately 9.8 m/s².
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A student is planning on making a change to a circut witch changhe will increase the current
There are several ways to increase the current in an electrical circuit, but it's important to keep in mind that any changes made should be done safely and within the limitations of the circuit components. Here are some possible options:
Decrease resistance: Ohm's Law states that current is directly proportional to voltage and inversely proportional to resistance. Therefore, decreasing resistance in a circuit will increase the current flowing through it. This can be done by replacing a high-resistance component with a lower-resistance one.
Increase voltage: Similarly, increasing the voltage in a circuit will also increase the current, as long as the resistance remains constant. This can be done by connecting a higher voltage power supply or battery to the circuit.
Add parallel branches: Adding more branches to a circuit in parallel can increase the overall current, as each branch provides an additional path for current to flow.
Increase capacitance or inductance: In circuits that contain capacitors or inductors, increasing the capacitance or inductance can increase the amount of current flowing through the circuit.
It's important to note that any changes made to a circuit should be done carefully and with an understanding of the potential risks involved. Seek guidance from a qualified professional if necessary, and always follow proper safety protocols.
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There is a 1.5 kg box on the table.
a) Draw the forces acting on the box and the table.
b) Calculate the weight of the box, its weight and the counteraction force of the table.
Answer:
a) The forces acting on the box and the table are as follows:
Weight of the box acting downwards (due to gravity)
Normal force acting upwards (from the table)
Frictional force (if the box is not moving, the frictional force will be equal and opposite to any force applied to move it)
b) The weight of the box is given by:
Weight = Mass x Acceleration due to gravity
Weight = 1.5 kg x 9.81 m/s^2
Weight = 14.715 N
The weight acts downwards due to gravity.
The counteraction force of the table is equal in magnitude and opposite in direction to the weight of the box, as per Newton's Third Law of Motion.
Therefore, the counteraction force of the table on the box is also 14.715 N, but it acts upwards to balance the weight of the box.
Explanation:
what are someone of the sources of errors in the ohm's law experiment
Explain what types of data streams can support and how they handle the data.
What are they?
Data streams are continuous flows of data that can be used to capture, process, and analyze real-time information.
Types of data streams that can be supported include:Data streams are typically handled by streaming data processing engines. These engines process and analyze the data as soon as it arrives, allowing for real-time insights and decision-making.
Sensor data streams: This type of data stream captures data from various sensors, such as temperature, humidity, motion, and pressure.Web service data streams: These data streams capture data from web services, such as weather, traffic, and stock market information.Database data streams: These data streams capture data from databases, such as customer data, product information, and financial transactions.Social media data streams: These data streams capture data from social media sites.Machine data streams: These data streams capture data from machines, such as production lines, robots, and industrial equipment.They are used to capture data from a variety of sources, such as sensors, web services, databases, and other online sources.
Data streams are handled in real-time, meaning that they are processed and analyzed as soon as they arrive.
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water is sprayed on oranges during a frosty night. if an average of 11.8g of water freezes on each orange, how much heat is released answer
Answer:
fusion of water = 80 cal/gm
11.8 g * 80 cal / g = 940 cal released by each orange
an 8-meter ladder is leaning against a vertical wall. if a person pulls the base of the ladder away from the wall at the rate of 0.7 m/s, what is the rate of change of the distance between the top of the ladder and the ground when the base of the ladder is
The rate of change of the distance between the top of the ladder and the ground when the base of the ladder is pulled away from the wall at the rate of 0.7 m/s is: 0.7 m/s.
Since the base of the ladder is moving away from the wall, the distance between the top of the ladder and the ground must be increasing. Therefore, the rate of change of the distance between the top of the ladder and the ground is equal to the rate of change of the base, which is 0.7 m/s.
The formula for the rate of change is given by: Rate of change = Change in distance/Change in time.
In this case, the rate of change is equal to the rate at which the base is moving away from the wall, 0.7 m/s.
To summarize, the rate of change of the distance between the top of the ladder and the ground when the base of the ladder is pulled away from the wall at the rate of 0.7 m/s is 0.7 m/s.
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What happens to light when it travels from air into water?
If the skid mark from a crashed vehicle measures approximately 70 ft, and the measured friction coefficient is 0.25, what was the initial vehicle speed prior to braking?
A. 65 mph
B. 40 mph
C. 23 mph
D. 15 mph
Therefore, the answer is C. 23 mph is the maximum initial speed that would result in a skid mark of 70 ft with a friction coefficient of 0.25.
What is the formula for final velocity?Initial velocity (u) squared plus two times the acceleration (a) times the displacement equals final velocity (v) squared (s). Final velocity (v) is equal to the square root of initial velocity (u) squared plus two times the acceleration (a) times displacement when v is the variable being solved for (s).
[tex]v^2 = u^2 + 2as[/tex]
Where:
v is the final velocity (0 mph, as the vehicle comes to a stop)
u is the initial velocity we want to find
a is the acceleration due to friction, which is equal to the friction coefficient multiplied by the acceleration due to gravity [tex](a = 0.25 * 32.2 ft/s^2 = 8.05 ft/s^2)[/tex]
s is the distance the vehicle traveled before coming to a stop, which is the length of the skid mark (s = 70 ft)
Substituting the given values into the equation, we get:
[tex]0^2 = u^2 + 2(0.25 * 32.2 ft/s^2)(70 ft)\\[/tex]
Simplifying:
[tex]0 = u^2 + 2(8.05 ft/s^2)(70 ft)\\0 = u^2 + 1134 ft^2/s^2\\u^2 = -1134 ft^2/s^2[/tex]
Since we cannot have a negative value for velocity, we can conclude that the initial velocity must be less than 23 mph.
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Aaron, Bonny and Connor are playing a game with cards.
Aaron has some cards.
Bonny has twice as many cards as Aaron.
Connor has 6 cards more than Bonny.
They have a total of 101 cards.
Work out how many cards Aaron has.
The number of cards Aaron had are 19 when Aaron, Bonny and Connor are playing a game with cards which are a total of 101.
Given that Aaron, Bonny and Connor are playing a game with cards.
Let the number of cards Aaron had = x
Bonny has twice the cards of Aaron = 2x
Number of cards Connor had are 6 more than Bonny = 2x + 6
The total number of cards = 101
Then by adding the number of cards that Aaron, Bonny and Connor had we get the total number of cards.
Then Aaron cards + Bonny cards + Connor cards = 101
x + 2x + 2x + 6 = 101
5x = 101 - 6 = 95
x = 95/5 = 19
Hence the total number of cards Aaron had = 19
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in one to two sentences explain how electromagnets move a maglev train
Electromagnets are used to create a magnetic field which levitates the train above the track, and then by varying the current in the electromagnets, the train can be propelled forward or slowed down or stopped.
How does magnetic train levitate?Maglev trains use the principle of electromagnetic suspension to levitate and propel the train. Electromagnets are placed on the underside of the train, and these electromagnets are energized with a current that creates a magnetic field.
This magnetic field interacts with the magnetic field of a conductor in the track, which creates an upward force that levitates the train above the track.
Once the train is levitated, the next step is to propel it forward. The same electromagnets that are used for levitation can be used to propel the train forward.
By varying the current in the electromagnets, the magnetic field created can be made to push or pull the train forward or backward. This process is known as magnetic propulsion or maglev propulsion.
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g a truck with a mass of 1690 kg and moving with a speed of 13.0 m/s rear-ends a 615 kg car stopped at an intersection. the collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. find the speed of both vehicles after the collision in meters per second
The final velocities of the truck and car after the collision are both 13.0 m/s and 0 m/s, respectively.
Steps
To solve this problem, we can use the conservation of momentum and kinetic energy:
Conservation of momentum:
m1v1 + m2v2 = m1v1' + m2v2'
where
m1 = 1690 kg (mass of the truck)
v1 = 13.0 m/s (initial velocity of the truck)
m2 = 615 kg (mass of the car)
v2 = 0 m/s (initial velocity of the car)
v1' = final velocity of the truck
v2' = final velocity of the car
Solving for v1' and v2':
m1v1 + m2v2 = m1v1' + m2v2'
1690 kg * 13.0 m/s + 615 kg * 0 m/s = 1690 kg * v1' + 615 kg * v2'
21970 kg m/s = 1690 kg * v1' + 0 kg m/s
v1' = 21970 kg m/s / 1690 kg = 13.0 m/s
So the truck maintains its initial speed of 13.0 m/s after the collision.
Now let's solve for the final velocity of the car:
m1v1 + m2v2 = m1v1' + m2v2'
1690 kg * 13.0 m/s + 615 kg * 0 m/s = 1690 kg * 13.0 m/s + 615 kg * v2'
0 kg m/s = 0 kg m/s + 615 kg * v2'
v2' = 0 m/s
So the car comes to a complete stop after the collision.
Therefore, the final velocities of the truck and car after the collision are both 13.0 m/s and 0 m/s, respectively.
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brandon is on one side of a river that is 80 m wide and wants to reach a point 300 m downstream on the opposite side as quickly as possible by swimming diagonally across the river and then running the rest of the way. find the minimum amount of time if brandon can swim at 2 m/s and run at 4 m/s.
The minimum time Brandon can swim at 2 m/s is 144.5 seconds and run at 4 m/s is 55 seconds, with the distance traveled diagonally being 289 m
We have to find the minimum amount of time taken to reach the other side, as Brandon wants to reach a point 300 m downstream on the opposite side as quickly as possible by swimming diagonally across the river and then running the rest of the way.
So, the distance he covers in water = distance diagonally in water
And, the time he takes to cover this distance in water = distance/Speed => t = diagonal distance/speed of swimming
Also, the distance he covers in running = perpendicular distance
And, the time he takes to cover this distance in running = distance/Speed => t = perpendicular distance/speed of running
Given:
Width of river = AB = 80 m
Distance to reach = AC = 300 m
Speed of Swimming = 2 m/s
Speed of Running = 4 m/s
The minimum time taken will be the time taken to cover the total distance i.e. AB + BC
We know that AC = AB² + BC² (by Pythagoras theorem)
We have AC = 300, AB = 80, and speed of swimming = 2 m/s. Let BC = x, then we have:
x² = √(300² - 80²)
x² = √(90.000 - 6400)
x² = √(83.600)
x = 289 m
So, the time he takes to cover this distance in water is:
t = diagonal distance/speed of swimming
t = BD/Speed of swimming
t = 289/2 = 144.5
Now, he covers the remaining perpendicular distance CD by running which is equal to 220m
The time he takes to cover this distance in running is:
t = perpendicular distance/speed of running
t = CD/Speed of running
t = 220/4 = 55 s
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An object with mass m is attached to the end of a spring with spring constant k, the object is displaced a distance d from equilibrium and released.
What is the speed v of the mass when it returns to the equilibrium position?
When an object with mass m is attached to the end of a spring with spring constant k and displaced a distance d from equilibrium and released, it undergoes simple harmonic motion. The speed v of the mass when it returns to the equilibrium position is given by: v = sqrt((kd²)/m)
The maximum potential energy of the system is given by:
U = (1/2)kx²
where x is the displacement of the object from its equilibrium position. At the maximum displacement d, the potential energy of the system is:
U = (1/2)kd²
As the object oscillates back and forth, the potential energy is converted into kinetic energy, given by:
K = (1/2)mv²
where v is the speed of the mass at any point during its motion.
At the equilibrium position, all of the potential energy has been converted into kinetic energy, so we can equate the two expressions:
(1/2)kd² = (1/2)mv²
Solving for v, we get:
v = sqrt((kd²)/m)
Therefore, the speed v of the mass when it returns to the equilibrium position is given by:
v = sqrt((kd²)/m)
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a solenoid with 385 turns per meter and a diameter of 17 cm has a magnetic flux through its core of magnitude 1.28 x 10-4 t m2. (a) find the current in this solenoid (b) what happens if you double the diameter of the solenoid?
(a) The current in the solenoid is 0.85 A (b) If you double the diameter of the solenoid, the the magnetic flux through the solenoid will increase by 3.19 times. This is because on doubling the diameter, the area of the core will be quadrupled.
(a) The magnetic flux density B inside a solenoid is calculated using the following formula:
B = μ₀nI
Where, B is the magnetic flux density in teslas (T), μ₀ is the permeability of free space = 4π x 10⁻⁷ Tm/A, n is the number of turns per unit length of the solenoid, I is the current in amperes (A)
From the above equation, we can write
I = B/μ₀n
So, the current in the solenoid is given by
I = 1.28 x 10⁻⁴ / (4π x 10⁻⁷ × 385) = 0.85 A
(b) The magnetic flux through a solenoid is given by
Φ = BA
where, A is the cross-sectional area of the solenoid (area of a circle of diameter 17 cm) = πr² = π(17/2)² = 226 cm² = 0.0226 m², B is the magnetic flux density inside the solenoid as calculated above, Φ = 0.85 × 0.0226 = 0.0193 Wb
If we double the diameter of the solenoid, then the cross-sectional area A of the solenoid will be quadrupled because A ∝ d² (where d is the diameter of the solenoid).So, the new area of the solenoid will be
A = π(2r)² = π(2 × 17/2)² = 722 cm² = 0.0722 m²
So, the new magnetic flux through the solenoid will be
Φ' = Φ × (A'/A)Φ' = 0.0193 × (0.0722/0.0226) = 0.0615 Wb
Therefore, if we double the diameter of the solenoid, then the magnetic flux through the solenoid will increase by a factor of (0.0615/0.0193) = 3.19 times.
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A baseball is dropped from the top of a 85 m tall building. Ignoring air resistance, how fast will it hit the ground?
Answer:
40.8 m/s
Explanation:
We can use the kinematic equation that relates the final velocity (Vf) of an object dropped from rest to the distance it has fallen (h) and the acceleration due to gravity (g):
Vf^2 = 2gh
where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Plugging in the given values, we get:
Vf^2 = 2 * 9.8 m/s^2 * 85 m
Vf^2 = 1666
Vf = sqrt(1666) ≈ 40.8 m/s
Therefore, the baseball will hit the ground with a speed of approximately 40.8 m/s.
Convert the following: (a) 1 000 kg into g. (b) 1 000 000 m into km (c) 0.0000037 kg to mg. (d) 0.00000125 m to mm.
how do you do it
Answer:
a) 1000000 g
b) 1000 km
c) 3,7 mg
d) 0,00125 mm
Explanation:
a) 1kg has 1000g, so we can make a proportion:
1 kg - 1000 g
1000 kg - x g
Use the property of the proportion to find x:
[tex]x = 1000 \times 1000 = 1 \times {10}^{6} \: g = 1000000 \: g[/tex]
b) 1 km has 1000 m, so let's do the same thing as we did earlier:
1 km - 1000 m
x km - 1000000 m
[tex]x = \frac{1000000}{1000} = 1000 \: km[/tex]
c) 1 g has 1000 mg and 1 kg has 1000 g, so 1 kg will have 1000 × 1000 =
[tex] {10}^{ 6} \: mg = 1000000 \: mg[/tex]
1 kg - 1000000 mg
0,0000037 kg - x mg
x = 0,0000037 × 1000000 = 3,7 mg
d) 1 m has 100 cm and 1 cm has 10 mm, so that means 1 m has 100 × 10 = 1000 mm
1 m - 1000 mm
0,00000125 m - x mm
x = 0,00000125 × 1000 = 0,00125 mm
I think I got it right
you are running around check 5 km an hour and then you increase your speed to 10 km an hour by what factors did you increase your kinetic
When a student runs at a speed of 5 km/h and then increases his or her speed to 10 km/h, the kinetic energy of the student increases by a factor of four. This is because kinetic energy is proportional to the square of the speed.
The formula for kinetic energy is:
K = 1/2mv²
where K is the kinetic energy, m is the mass of the object and v is the speed.
Since the mass of the student does not change, we can calculate the ratio of kinetic energy at the two speeds as follows:
K₁/K₂ = (1/2)m(v₁²/v₂²)
where v₁ is the initial speed of 5 km/h and v₂ is the final speed of 10 km/h.
K₁/K₂ = (1/2)m(5²/10²) = (1/2)m(1/4) = (1/8)K₂
Therefore, the kinetic energy at the final speed of 10 km/h is four times greater than the kinetic energy at the initial speed of 5 km/h. This means that the student increased their kinetic energy by a factor of four.
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