A loop of conductor is at rest on a table. A magnet is brought nearby directly above the loop. What is the direction of current flow (looking from above) in the following situations?

a) The north pole of the magnet points down and the magnet is moved down

clockwise

counterclockwise

no current flows
b) The north pole of the magnet points up and the table is raised

clockwise

counterclockwise

no current flow
c) The loop of wire is rotated clockwise about its center axis

clockwise

counterclockwise

no current flow

Answers

Answer 1

Answer:

In each of the given situations, the direction of current flow in the loop of conductor will depend on the relative motion of the magnet and the loop, as well as the orientation of the magnet.

a) In the first situation, the magnet is brought down directly above the loop, with the north pole pointing down. As the magnet is moved down, the magnetic field produced by the magnet will pass through the loop of conductor and generate a current in the loop. The direction of this current will depend on the orientation of the loop relative to the magnet. If the loop is oriented such that the current flows in a clockwise direction when viewed from above, the current will be in the clockwise direction. If the loop is oriented such that the current flows in a counterclockwise direction when viewed from above, the current will be in the counterclockwise direction.

b) In the second situation, the magnet is brought near the loop, with the north pole pointing up. The table is then raised, causing the magnet to move relative to the loop. As the magnet moves, it will generate a changing magnetic field that will induce a current in the loop. The direction of this current will depend on the orientation of the loop relative to the magnet, as well as the direction of the magnet's movement. If the loop is oriented such that the current flows in a clockwise direction when viewed from above, and the magnet is moving upward, the current will be in the clockwise direction. If the loop is oriented such that the current flows in a counterclockwise direction when viewed from above, and the magnet is moving upward, the current will be in the counterclockwise direction.

c) In the third situation, the loop of wire is rotated clockwise about its center axis. This will not produce a current in the loop, as there is no relative motion between the loop and the magnet, and therefore no changing magnetic field to induce a current. Therefore, in this situation, no current will flow in the loop.


Related Questions

in fig. 11-37, a small, solid, uniform ball is to be shot from point p so that it rolls smoothly along a horizontal path, up along a ramp, and onto a plateau.then it leaves the plateau horizontally to land on a game board, at a horizontal distance d from the right edge of the plateau. the vertical heights are h1 ! 5.00 cm and h2 ! 1.60 cm. with what speed must the ball be shot at point p for it to land at d ! 6.00 cm?

Answers

The speed at which the ball must be shot at the point p for it to land at the distance 6 cm is 6.635 m/s.

Given that,

Height h₁ = 5 cm

Height h₂ = 1.6 cm

Distance d = 6 cm

We know, t = √(2 h₂/g) = √(2* 1.6)/9.8 = 0.57 s

We know that, distance is nothing but speed multiplied by time.

According to the above statement, expression can be written as d = V₁* t

V₁ = d/t = 6/0.57 = 10.52 m/s

Using work energy theorem, we have

mg(h₁ - h₂) = 1/2* m * V₁² - 1/2* m * V₂²

Eliminating 'm' on both sides, g(h₁ - h₂) = 1/2* V₁² - 1/2* V₂²

9.8 (5 - 1.6) = 1/2* 10.52² - 1/2* V₂²

33.32 = 1/2(10.52² - V₂²)

(10.52² - V₂²) = 66.64

V₂² = 44.03

V₂ = 6.635 m/s

Thus, the speed of the ball with which it should be shot is 6.635 m/s.

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taipei 101 (a 101-story building in taiwan) is sited in an area that is prone to earthquakes and typhoons, both of which can lead to dangerous oscillations of the building. to reduce the maximum amplitude, the building has a tuned mass damper, a 660,000 kg mass suspended from 42-m-long cables that oscillates at the same natural frequency as the building (figure 1). when the building sways, the pendulum swings, reaching an amplitude of 75 cm in strong winds or tremors. damping the motion of the mass reduces the maximum amplitude of oscillation of the building.

Answers

The period of oscillation of the building will be=13 s

During strong winds, the pendulum moves when it passes through the equilibrium position with a speed of = 0.36 m/s

(a)

Let us assume that the time period of oscillations of the building be [i 0]

So, the expression for time period will be: [tex]$$T=2 \pi \sqrt{\frac{l}{g}}$$[/tex]

Where, l is the length of the pendulum and g is the acceleration due to gravity.

So, the value of time period of oscillations will be:

[tex]T & =2 \pi \sqrt{\frac{(42 \mathrm{~m})}{\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)}}[/tex] =13.0 seconds.

(b)

As we have, the relation between angular frequency and time period, which is [tex]$$\omega=\frac{2 \pi}{T}$$[/tex]

So, the value of the linear velocity of the pendulum will be

[tex]$$\begin{aligned}v & =A \omega \\& =A\left(\frac{2 \pi}{T}\right) \\& =\frac{2 \pi A}{T} \\& =\frac{2 \pi(0.75 \mathrm{~m})}{(13.0 \mathrm{~s})} \\& =0.36 \mathrm{~m} / \mathrm{s}\end{aligned}$$[/tex]

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The complete question should be:

Taipei 101 (a 101-story building in Taiwan) is sited in an area that is prone to earthquakes and typhoons, both of which can lead to dangerous oscillations of the building. To reduce the maximum amplitude, the building has a tuned mass damper, a 660,000 kg mass suspended from 42-m-long cables that oscillates at the same natural frequency as the building. When the building sways, the pendulum swings, reaching an amplitude of 75 cm in strong winds or tremors. Damping the motion of the mass reduces the maximum amplitude of oscillation of the building.

What is the period of oscillation of the building?

During strong winds, how fast is the pendulum moving when it passes through the equilibrium position?

A 10.0-kg microwave oven is pushed 8.00 m up the sloping surface of a loading ramp inclined at an angle of above the horizontal, by a constant force \vec{F} F with a magnitude 110 N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250. (a) What is the work done on the oven by the force ? (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven’s kinetic energy. (e) Use \Sigma \vec{\boldsymbol{F}}=m \vec{\boldsymbol{a}}Σ F =m a to calculate the acceleration of the oven. Assuming that the oven is initially at rest, use the acceleration to calculate the oven’s speed after traveling 8.00 m. From this, compute the increase in the oven’s kinetic energy, and compare it to the answer you got in part (d)

Answers

(a) Work done on the oven = 110J

(b) Work done by the friction force =250J

(c) Increase in PE =1000J

(d) Increase in KE = 640J

(e) Acceleration of the oven =14.9 m/s² And KE =74.5J

Given

g = 10

mass = 10kg

distance = 8 m

Magnitude of force = 110 N

Kinetic friction = 0.250 N

A

The work done on the oven by the force is given by

W = F.d

W = 110*14cos(0)

It is a Horizontal component

W = 110 J

B

Now to calculate the work done by the friction force

Frictional force = μ * N

Frictional force = 0.25*10*10

Frictional force = 25 N

Frictional work = frictional force . d

work(f) = 25 * 10cos(0)

Work(f) = 250 J

C

Increase in the Potential  energy

PE = mgh

PE = 10*10*10cos(0)

PE = 1000 J

D

The increase in the oven's kinetic energy

PE₁ = KE₁ + W = PE₂ + KE₂

110 - 260 - 1000 = KE₂

KE₂ = 640J

E

The acceleration of the oven

Force - frictional force = ma

a = 110 -250/10 = 14 m/s²

Now to calculate KE we should know the velocity

V(f)² = v(i)² + 2ad

v(f)² = 2(14)(8) = 224

v(f) = 14.9 m/s

Now KE will be given as

(1/2)mv²

KE = (1/2)(10)(14.9)

KE  = 74.5 J

Therefore, we got to know the work done, PE and KE just by the force applied.

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under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. when the drive force is shut off, the snowmobile coasts to a halt. the snowmobile and its rider have a mass of 136 kg. under the influence of a drive force of 205 n, it is moving at a constant velocity whose magnitude is 5.50 m/s. the drive force is then shut off.Find (a) the distance in which the snowmobilecoasts to a halt and (b) the time required to dos

Answers

There is no difference in the shape of 2p and 3p orbitals since the azimuthal quantum number, which defines the shape of the orbital, is the same for both.

The electron orbital energy levels are listed in the following order, from lowest to highest: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p. Because electrons all have the same charge, they try to stay as far apart as possible due to repulsion. We only need three hybrid orbitals, aka sp2, because the carbon has no lone pairs and is connected to three hydrogens. Don't forget to account for any lone pairings. Every solitary pair need its own hybrid orbital.

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an ellipse is the locus of points in a plane such that the sum of the distances from any point in the locus to two points, called the foci, is a .

Answers

An ellipse is the location of all the points in a plane whose distances from two fixed points in the plane add up to a constant value. The foci, or singular focus, are the fixed points that are encircled by the curve.

what is an ellipse?

In mathematics, an ellipse is the location of points in a plane so that their separation from a fixed point has a fixed ratio of "e" to their separation from a fixed line (less than 1). The conic section, which is the intersection of a cone with a plane that does not intersect the base of the cone, includes the ellipse. The fixed point is called the focus and is denoted by S, the constant ratio 'e' as the eccentricity, and the fixed line is called as directrix (d) of the ellipse.

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which of the following situations is an example of a negative feedback loop? group of answer choices a thermostat regulating the temperature of your home rolling a ball of snow down a hill the greenhouse effect a viral video on the internet

Answers

The greenhouse effect is one instance of a negative feedback loop among the examples presented. There are several instances of positive feedback loops.

Positive feedback loop: what is it?

Positive feedback, also known as aggravating feedback or self-reinforcing feedback, is a phenomenon that takes place in a feedback loop that amplifies the impact of a little disruption. In other words, a perturbation's impacts on a system include an increase in the perturbation's size.

A negative feedback loop is what?

When a system, process, or mechanism's output is fed back in a way that seeks to lessen oscillations in the output, whether brought on by changes in the input or other disturbances, this is known as negative feedback (or balancing feedback).

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quasars may result from the very energetic merging of in condensing galaxies. quasars may result from the very energetic merging of in condensing galaxies. black holes suns stars dark matter

Answers

Quasars may result from the very energetic merging of dark matter in condensing galaxies.

Quasars may result from the very energetic merging of gas spiraling at high velocity in condensing galaxies.

What is  a quasar?

A quasar forms can be regarded as one that the material that falls into the accretion disc  which is seen in  supermassive black hole at the center of a galaxy.

A massive object, like a galaxy cluster, is able to deform the space-time shape as a consequence of its own gravity, so the light that it is coming from a source that is behind it in the line of sight will be bend or distorts in a way that will be magnified, making small arcs around the cluster with the image of the background object.

Not only does the central engine of active galaxies and quasars require a black hole, Accretion disk of matter is also needed to provide the energy radiated. Dust, other stellar debris, and gas that has come near to a black hole but fallen into it yet forms a flattened band of spinning matter around the event horizon called the accretion disk (or disc).

Therefore, quasars may result from the very energetic merging of dark matter in condensing galaxies.

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Quasars may result from the very energetic merging of dark matter in condensing galaxies.

Quasars may result from the very energetic merging of gas spiraling at high velocity in condensing galaxies.

What is  a Quasar?

A quasar forms can be regarded as one that the material that falls into the accretion disc  which is seen in  supermassive black hole at the center of a galaxy.

A massive object, like a galaxy cluster, is able to deform the space-time shape as a consequence of its own gravity, so the light that it is coming from a source that is behind it in the line of sight will be bend or distorts in a way that will be magnified, making small arcs around the cluster with the image of the background object.

Not only does the central engine of active galaxies and quasars require a black hole, Accretion disk of matter is also needed to provide the energy radiated. Dust, other stellar debris, and gas that has come near to a black hole but fallen into it yet forms a flattened band of spinning matter around the event horizon called the accretion disk (or disc).

Therefore, quasars may result from the very energetic merging of dark matter in condensing galaxies.

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A pulse of electromagnetic radiation is propagating in the +y direction. You have two devices that can detect electric and magnetic fields. You place detector #1 at location < 0, -3, 0> m and detector #2 at location < 0, 3, 0> m.
(a) At time t = 0, detector #1 detects an electric field in the -x direction. At that instant, what is the direction of the magnetic field at the location of detector #1?
(b) At what time will detector #2 detect electric and magnetic fields?

Answers

a. At the time [tex]\rm \(t = 0\)[/tex], the magnetic field at the location of detector #1 is in the +z direction.

b. At [tex]\(t \approx 2 \times 10^{-8}\)[/tex] seconds, detector #2 will detect the electric and magnetic fields.

(a) To determine the direction of the magnetic field at the location of detector #1, we can use the right-hand rule, which relates the direction of the magnetic field [tex]\rm (\(B\))[/tex], the direction of the electric field [tex]\rm (\(E\))[/tex], and the direction of propagation [tex]\rm (\(+y\))[/tex].

Since the pulse of electromagnetic radiation is propagating in the +y direction and detector #1 detects an electric field in the -x direction, the magnetic field [tex]\rm (\(B\))[/tex] must be in the +z direction.

This is because electromagnetic waves are transverse waves, and the electric and magnetic fields are perpendicular to each other and the direction of propagation.

So, at time [tex]\rm \(t = 0\)[/tex], the magnetic field at the location of detector #1 is in the +z direction.

(b) To determine the time at which detector #2 will detect electric and magnetic fields, we need to consider the distance between the two detectors and the speed of propagation of the electromagnetic wave.

Let's assume that the speed of light [tex]\rm (\(c\))[/tex] is [tex]\rm \(3.00 \times 10^8\)[/tex] m/s.

The distance between the two detectors is [tex]\rm \(d = 6\)[/tex] m (distance between the y-coordinates of the detectors).

The time [tex]\rm (\(t\))[/tex] it takes for the electromagnetic wave to travel from detector #1 to detector #2 is given by:

[tex]\[ t = \frac{d}{c} \]\\\\\ t = \frac{6 \, \text{m}}{3.00 \times 10^8 \, \text{m/s}} \]\\\\\ t \approx 2 \times 10^{-8} \, \text{s} \][/tex]

So, in [tex]\rm \(t \approx 2 \times 10^{-8}\)[/tex] seconds, detector #2 will detect the electric and magnetic fields.

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N А. В. S Current Current is moving through a copper ribbon in the direction shown. The ribbon is suspended between the poles of a magnet as indicated. When the voltage is measured between A and B, it is observed that A is at higher potential than B. Which of the following explains this observation? A The electrons in the ribbon behave like magnetic dipoles; thus, they align with the magnetic field of the magnet that deflects electrons in the ribbon toward point A. B The electrons in the ribbon behave like magnetic dipoles; thus, they align with the magnetic field of the magnet that deflects electrons in the ribbon toward point B. с Electrons moving through the magnetic field experience a magnetic force to the left that deflects electrons in the ribbon toward point A. D Electrons moving through the magnetic field experience a magnetic force to the right that deflects electrons in the ribbon toward point B. E The electric field in the ribbon is distorted by the magnetic field. This distortion creates an electric field component directed from point A to point B that deflects electrons in the ribbon toward point B.

Answers

The explanation for this observation is that the magnetic field distorts the electric field in the band. This distortion creates an electric field component directed from point A to point B which deflects the electrons in the bar toward point B.

The physical field that surrounds an electrically charged particle and exerts a force on all other charged particles in that field, either attracting or repelling them, is called the electric field (or E-field). It can also refer to the physical field system of charged particles.

The negatively charged electrons are directed to the positive electrode or (+) plate. Positively charged protons are directed to the negative electrode or plate (-).

Therefore, ferromagnetic materials bend the lines of magnetic force in space in a static field. The distortion effect on the static magnetic field is disturbed by the deformed material.

The complete questions:

Current is moving through a copper ribbon in the direction shown. The ribbon is suspended between the poles of a magnet as indicated. When the voltage is measured between A and B, it is observed that A is at higher potential than B. Which of the following explains this observation?

A) The electrons in the ribbon behave like magnetic dipoles, thus, they align with the magnetic field of the magnet that deflects electrons in the ribbon toward point A.

B) The electrons in the ribbon behave like magnetic dipoles, thus, they align with the magnetic field of the magnet that deflects electrons in the ribbon toward point B.

C) Electrons moving through the magnetic field experience a magnetic force to the left that deflects electrons in the ribbon toward point A

D) Electrons moving through the magnetic field experience a magnetic force to the right that deflects electrons in the ribbon toward point B

E) The electric field in the ribbon is distorted by the magnetic field. This distortion creates an electric field component directed from point A to point B that deflects electrons in the ribbon toward point B.

The true choice is E

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The rate per hundredweight is usually based on the shipment origin and destination, although the actual price charged for a particular shipment is normally subject to a minimum charge and may also be given a(n) ____
surcharge

Answers

The rate per hundredweight is usually based on the shipment origin and destination, although the actual price charged for a particular shipment is normally subject to a minimum charge and may also be given a(n) surcharge.

A surcharge is any additional fee, tax, or charge that is tacked onto the cost of a good or service above and above the initial price that was provided. The advertised price of the commodity or service does not include a surcharge, which is frequently added to an already-applied tax. The cost of a surcharge can range and be either a set sum or a percentage of A governing body may apply this fee in order to raise additional funds or cover the expense of rising commodity prices. Consumers who purchase specific goods and services may be subject to surcharges, which are additional fees and/or taxes. At the moment of sale, they are included in the cost of the purchase. Thus, whenever you buy something,

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At the time of its discovery, 3C 273 was the most ____________________ object in the universe.

Answers

3C 273 was the most luminous object in the universe at the discovery time.

What are stars?

An illuminated sphere of plasma that is held together by its gravity is a star.

The Sun is the star that is closest to Earth. Due to their enormous distance from Earth, many more stars are also visible at night to the unaided eye from Earth, appearing as a multiplicity of stationary light dots in the sky.

In the past, the brightest stars were assigned names and grouped into constellations and asterisms.

Star catalogs have been put out by astronomers that list the known stars and offer standardized stellar labels.

All stars outside of our galaxy, the Milky Way, and the majority of other stars in the universe are, nevertheless, invisible to the human eye from Earth.

Even the most powerful telescopes cannot see the majority of them from Earth.

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what must the emf e of the battery be in order for a current of 2.00 a to flow through the 5.00 v battery, as shown?

Answers

The battery's emf has to be -108.75 volts for a 2a current to pass through a 5.00 v battery.

Kirchhoff's Voltage Law states that for any closed network, the voltage around a loop is equal to the total of all voltage drops in that loop and is equal to zero. The conservation of energy, or to put it another way, is the characteristic of Kirchhoff's law which states that the algebraic sum of each voltage in the loop must equal zero.

Do a series/parallel reduction first. Apply Kirchhoff's laws now to get the answer.

ΔV adefa ​ =0 or − (20Ω) × (2A) × 5V − (20Ω) × I2 ​

= 0 or I2 ​

= −2.25A

I1 ​+ I2 ​= 2A or I1 ​= 2A − ( −2.25A ) = 4.25A

ΔV adefa ​= 0 or (15Ω) × (4.25A) + ϵ − (20Ω) (−2.25A) = 0

ϵ = −108.75V; Reversing polarity is necessary.

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Most of the spaceflight missions to the outer planets have been flyby missions, but two of them were orbiters. Which of these two spacecraft orbited giant planets? Check the correct two.
(Hint: An orbiter might do a flyby of one planet and then go on to orbit a differentplanet.)
Group of answer choices
Galileo
Ulysses
New Horizons
Voyager
Cassini
Pioneer

Answers

Galileo and Cassini are the two missions on outer planets that were orbiters. Galileo orbited Jupiter for eight years, while Cassini orbited Saturn for thirteen years.

A total of 9 spacecraft have been launched for missions on outer planets. Two of them are orbiters. Galileo was an American spacecraft to study Jupiter. It was delivered into Earth orbit on October 18, 1989, by Space Shuttle Atlantis,

Cassini was a space mission by NASA and ASI to study the planet Saturn. It was launched on Titan IVB/Centaur, and the date was October 15, 1997. Cassini was active for nearly 20 years which it spent orbiting Saturn for 13 years. It was the first spacecraft to enter Saturn orbit.

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Some electric-power companies use water to store energy. Water is pumped by reversible turbine pumps from a low to a high reservoir. To store the energy produced in 1 hour by a 120 MW electric-power plant, how many cubic meters of water will have to be pumped from the lower to the upper reservoir? Assume the upper reservoir is 470 m above the lower and we can neglect the small change in depths within each. Water has a mass of 1000 kg for every 1.0{m}^3.

Answers

The volume of water that will be pumped from the lower to the upper reservoir to generate the energy is 93,790.7 m³.

What is the volume of water needed to generate the power?

The volume of water needed to generate the given power is calculated by applying the law of conservation of energy as shown below.

gravitational potential energy of water or work done by the water = energy generated

Mathematically, the equation is given as;

W = PV

where;

W is work done by the waterV is the volume of the waterP is the pressure of the water

W = (ρgh)V

where;

ρ is the density of waterg is acceleration due to gravityh is the height through which the water is pumped

(ρgh)V  = Pt

where;

P is the power generatedt is the time in which the power was generated

V = Pt / ρgh

The given parameters include;

P, power = 120 MW = 120 x 10⁶ Wt , time = 1 hour = 3600 sρ, density of water, = 1000 kg/m³g, acceleration due to gravity = 9.8 m/s²h, height of water, = 470 m

The volume of water needed to generate the energy is calculated as follows;

V = (120 x 10⁶ x 3600) / (1000 x 9.8 x 470)

V = 93,790.7 m³

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through a refinery, fuel ethanol is flowing in a pipe at a velocity of 1 m/s and a pressure of 101300 pa. the refinery needs the ethanol to be at a pressure of 2 atm (202600 pa) on a lower level. how far must the pipe drop in height in order to achieve this pressure? assume the velocity does not change. (hint: use the bernoulli equation. the density of ethanol is 789 kg/m3 and gravity g is 9.8 m/s2. pay attention to units!)

Answers

The pipe must drop in height by 13.101 m in order to achieve this pressure.

Bernoulli stated that at whatever point along the fluid flow, the total amount of pressure, kinetic energy and potential energy per unit volume is the same.

P + 1/2 ρv² + ρ* g* h = constant

Given that, v = 1 m/s

P₁ = 101300 Pa

P₂ = 202600 Pa

ρ ethanol = 789 kg/m³

g = 9.8 m/s²

Now, let us write the bernoulli equation,

ρ*g*h₁ + P₁ = ρ*g*h₂ + P₂

P₁ - P₂ =  ρ*g*h₂ - ρ*g*h₁

( h₂ - h₁) = (P₁ - P₂)/ρ*g

Δh = (P₁ - P₂)/ρ*g

Δh = -101300 / ( 789 * 9.8) = -13.101 m

Minus sign indicates that height must be reduced.

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What happens to length of metals as the temperature increases?

Answers

When a metal is heated, its atoms begin to vibrate more vigorously and move further apart from each other. This increases the distance between the atoms, which in turn causes the metal to expand and become longer.

The amount of expansion that a metal undergoes when heated depends on several factors, including the type of metal, the initial temperature of the metal, and the rate at which it is heated. In general, most metals will expand by a small amount when heated over a wide range of temperatures.

At very high temperatures, some metals may begin to undergo a phase change, such as melting or vaporization. This can cause the metal to undergo a significant change in length, depending on the amount of phase change that occurs.

Overall, the length of a metal will generally increase as the temperature increases, although the exact amount of expansion will depend on the specific conditions and the type of metal involved.

on integrated circuits the dc biasing of bjts is usually done using resistors and capacitors, since transistor current sources are too expensive. true or false

Answers

Due to the high cost of transistor ecurrent sourcs, resistors and capacitors are typically used to bias bjts in integrated circuits.

A semiconductor wafer on which hundreds or millions of small resistors, capacitors, diodes, and transistors are manufactured is known as an integrated circuit (IC), also known as a chip, microchip, or microelectronic circuit. Integrated circuits, which can be classified as analogue, digital, or a hybrid of the two, are extensively utilised in electronics design today. Amplifiers, video processors, computer memory, switches, and microprocessors are just a few applications for ICs.

On a silicon wafer, integrated circuits are made up of reduced versions of transistors, microprocessors, and diodes.

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What is the wave phenomena that occurs whey a wave passes you and the pitch seems to change?

Answers

Answer:

The wave phenomenon that you are describing is known as the Doppler effect. The Doppler effect occurs when a wave, such as a sound wave, passes by an observer and the pitch of the wave seems to change. This is because the frequency of the wave changes as it moves closer to or farther away from the observer. The pitch of a sound wave is directly related to its frequency, so when the frequency of the wave changes, the pitch will also change. This effect is commonly observed with sound waves, but it can also occur with other types of waves, such as light waves.

a 355 ml soda can is 6.2 cm in diameter and has a mass of 20 g. such a soda can half full of water is floating upright in water. What length of the can is above the water level?

Answers

The length of the can above the water level is 5.22 cm.

First determine the height of the can.

355cm3 = pi * r2 * l     Volume of cylinder formula where l is length of can.

355=pi * (3.1)2 * l       Where 3.1 is half of the diameter.

l = 11.76cm

Now determine the total mass of the can including the water inside.

Total mass = 20g + (355ml)/2 * 1g/ml = 177.5g,    where 1g/ml is the density of water, therefore since we know half the can is full of water we take the volume of water inside, multiply it by the density and determine the mass of water. This, in addition to the weight of the can itself gives us the total mass of the floating object.

Total mass=197.5g

According to Archimedes' principle the mass of a floating object equals the mass of fluid displaced by the object.

197.5=mass of water displaced

197.5=density of water * volume of water

197.5= 1g/ml * (pi * (3.1)2 * l)      where l is the length of can inside the water

6.54cm=l

So take total length of can and subtract the submerged length to retrieve the above water length.

11.76cm - 6.54cm = 5.22cm

5.22 cm of water level.

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Yvette does not understand why her plant is not growing. Through observation of the care she gave it, she decided, if I increase the water from 1 cup a week to 2 cups a week, then it will grow. What has she just proposed?

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Answer:

Yvette has just proposed a hypothesis. A hypothesis is a proposed explanation for a phenomenon or observation. In this case, Yvette has observed that her plant is not growing and has proposed that increasing the amount of water it receives will cause it to grow. To test this hypothesis, Yvette could try increasing the amount of water the plant receives and observe whether it starts to grow. If the plant does start to grow, this would provide support for Yvette's hypothesis. However, if the plant does not start to grow, this would suggest that the hypothesis is incorrect and Yvette would need to come up with a new explanation for why the plant is not growing.

Before quarks were proposed, this scientist proposed partons to analyze high-energy hadroncollisions. Using a formalism fully developed by this person, the quantum mechanical amplitude canbe found by integrating over all the possible paths with a weighting factor given by the classicalaction, the so-called path integral formulation.

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Richard Feynman was the scientist who first suggested partons to study high-energy hadron collisions before quarks were suggested.

Quarks are what?

The elementary particle known as the quark is a fundamental component of matter. Hadrons, which are composite particles made of these quarks and neutrons and protons, the building blocks of atomic nuclei, are the most stable of these hadrons.

How do hadrons work?

A strong contact between two or more quarks holds together hadrons, which are composite subatomic particles. Simply put, hadrons are like molecules that the electric force holds together. Protons and neutrons combine to form hadrons, which are particles.

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a vechile ravels at a constant speed of 6.0 meters per second round a horrizontal circular curve with a radius of 24 meters. the mass of the vechile is 4.4 x 10^3 kolgranms, an icy patch is location at p on the curve

Answers

17) The magnitude of the frictional force that keeps the vehicle on its circular path is  C) 6.6 x 103 N

The magnitude of the frictional force that keeps the vehicle on its circular path can be found by using the equation For = mv2/r, where m is the mass of the vehicle, v is the velocity of the vehicle, and r is the radius of the curve. Plugging in the given values, we get Ffr = (4.4 x 103 kg)(6.0 m/s)2/24 m = 6.6 x 103 N.

18) The amount of work done on the object is D) 30 J.

Work is equal to the amount of force multiplied by the distance the object has moved. In this case, the force is 5 N and the distance moved is 3 m, so the work done on the object is 5 N x 3 m = 15 Nm = 15 J.

19) The crane does work at the rate of C) 5 x 104 watts

The rate at which work is done is equal to the total amount of work done divided by the total time it took to do the work. In this case, the total amount of work done is 200 N x 50 m = 10,000 Nm = 10,000 J, and the total time taken is 5 s. So the rate of work done is 10,000 J/5 s = 2,000 J/s = 2 kW = 5 x 104 watts.

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Complete question:

A vehicle travels at a constant speed of 6.0 meters per second around a horizontal circular curve with a radius of 24 meters. The mass of the vehicle is 4.4 × 103 kilograms. An icy patch is located at P on the curve.

17) What is the magnitude of the frictional force that keeps the vehicle on its circular path?

A) 6.5 × 104 N B) 1.1 × 103 N C) 6.6 × 103 N D) 4.3 × 104 N

18) A net force of 5.0 newtons moves an object a distance of 3.0 meters in. How much work is done on the object?

A) 10. J B) 15 J C) 1.0 J D) 30. J

19) A crane raises a 200‐newton weight to a height of 50 meters in 5 seconds. The crane does work at the rate of

A) 2 × 103 watts B) 8 × 10−1 watt C) 5 × 104 watts D) 2 × 101 watts

A boy swings a rubber ball attached to a string over his head in a horizontal, circular path. The piece of string is 0.735 m long and the ball
makes 159 complete turns each minute. What is the ball's centripetal acceleration?

Answers

Answer:

approximately 1.007 * 10^8 m/s^2

Explanation:

To find the ball's centripetal acceleration, we can use the formula a = v^2 / r, where a is the centripetal acceleration, v is the tangential velocity of the ball (the speed at which it moves around the circle), and r is the radius of the circle (the length of the piece of string).

First, we need to find the tangential velocity of the ball. The ball makes 159 complete turns each minute, which means it makes 159 * 2 * pi = 998.46 radians per minute. Since the ball is moving at a constant speed, the tangential velocity is equal to the angle it travels in radians per unit of time, so the tangential velocity of the ball is 998.46 radians per minute.

Next, we need to find the radius of the circle. The length of the piece of string is 0.735 m, and since the ball is swinging in a horizontal circle, the radius of the circle is equal to the length of the piece of string. Therefore, the radius of the circle is 0.735 m.

Now that we have the tangential velocity and the radius of the circle, we can plug these values into the formula to find the centripetal acceleration:

a = v^2 / r

= 998.46^2 / 0.735

= 1.007 * 10^8 m/s^2

Therefore, the ball's centripetal acceleration is approximately 1.007 * 10^8 m/s^2.

Show schematically and discuss how several semiconductor materials might be used together to obtain a more efficient solar cell.

Answers

A semiconductor material is a kind of electronic material they have semiconductor properties and can be used to make a different type of semiconductor devices and integrated circuits.

Other external factors such as light, heat, magnetism, and electricity will act on semiconductors and raise some physical effects and phenomena, which can be referred to as the semiconductor properties. Most of the base materials constituting solid-state electronic devices are semiconductors.

Different types of semiconductor devices have different types of functions and characteristics because of the various semiconductor properties.

Silicon is, by far, the most commonly used semiconductor material used in solar cells, representing approximately 95% of the modules sold today. It is also the second most abundant material on Earth (after oxygen) and the most common semiconductor used in computer and mobile chips.

Working of solar When light shines on a photovoltaic (PV) cell also called a solar cell that light may be reflected, absorbed, or pass right through the cell.

The photovoltaic cell is composed of semiconductor material; the “semi” means that it can conduct electricity better than an insulator but not as well as a good conductor like a metal. There are several different semiconductor materials used in PV cells.

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you are designing an elevator for a hospital. the force exerted on a passenger by the floor of the elevator is not to exceed 1.50 times the passenger's weight. the elevator accelerates upward with constant acceleration for a distance of 2.8 m and then starts to slow down. What is the maximum speed of the elevator?

Answers

The maximum speed of the elevator, the force exerted on a passenger by the floor of the elevator is not to exceed 1.50 times is 6m/s

ΣFy=may

n−mg=ma

n=1.5mg

a=0.60g=6ms^−2

v2=u2+2as=02+2×6×3

v=6m/s

The rate at which an object's distance traveled changes is measured by its speed. In terms of measurement, speed is a scalar, meaning it has magnitude but no direction. Speed is the rate at which an object moves over a given distance. a thing that travels at a high rate of speed and covers a lot of distance quickly. A slow-moving object, on the other hand, travels a comparatively short distance in the same amount of time when moving at a low speed. An object with zero speed is completely immobile.

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Select the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz 280 Hz b. 400 Hz 160 Hz 200 Hz e.180 Hz

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160 Hz is the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz.

According to this option c) is correct answer.

A string's fundamental frequency, succeeding frequencies, and two ends are defined as the sum of the entire depending on the number of harmonics. In other words, higher harmonics specify the frequencies that follow under the functions 2f, 3f, 4f, 5f, etc.

As a result, the higher harmonics would be:

1 x 80 Hz Equals 80 Hz (1st harmonic and Fundamental Frequency)

2 × 80Hz Equals 160Hz (2nd harmonic) (2nd harmonic)

3 x 80Hz Equals 240Hz (3rd harmonic)

4 × 80Hz = 320Hz (4th harmonic)

5 x 80Hz Equals 400Hz (5th harmonic)

As a result, the two higher harmonics of a string with an 80Hz fundamental frequency are 160Hz and 240Hz.

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I understand that the question you are looking for is:

Select the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz.

a)280 Hz

b) 400 Hz

c) 160 Hz

d) 200 Hz

e) 180 Hz

Electromagnetic radiation is moving to the right, and at this time and place the electric field is horizontal and points out of the page (see the figure). The magnitude of the electric field is E = 2600 N/C.
t is the magnitude of the associated magnetic field at this time and place?
B = __________T

Answers

The magnetic field at this time and place is 8.667 µT

Electromagnetic radiation consists of space-propagating electromagnetic field waves that carry momentum and electromagnetic radiation energy. These include radio waves, microwaves, infrared rays, light, ultraviolet rays, X-rays, and gamma rays.

Microwave is a form of electromagnetic radiation with wavelengths starting from approximately one meter to one millimeter corresponding to frequencies among 300 MHz and 300 GHz respectively. different assets define extraordinary frequency stages as microwaves.

Calculation:

Electric field E = 2600 N / C

Required magnetic field B = E / c

Where c = speed of EM radiation = 3 * 10 ^ 8 m / s

Plug the values weget B = 8.667* 10 ^ -6 T

= 8.667 µT

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a uniform disk with mass 43.6 kg and radius 0.300 m is pivoted at its center about a horizontal, frictionless axle that is stationary. the disk is initially at rest, and then a constant force 29.0 n is applied tangent to the rim of the disk.

Answers

The magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.320 revolutions will be 1.266 m/s

Mass = m = 43.6 Kg.

radius = 0.300 m

F = 29 n

So, value of Torque to the rim of disc will be = Fr = 30*0.3 = 9 Nm.

So, the value of angular acceleration will be: [tex]$\alpha=\frac{\tau}{I}: I=$[/tex] moment of inert of dice

[tex]=\frac{M R^2}{2}\\$\alpha=\frac{29 \mathrm{~N} \times 0.3 \mathrm{~m}}{\frac{43.6 \times \mathrm{kg} \times(0.3 \mathrm{~m})^2}{2}}\\\\=\frac{29 \times 0.3 \times 2}{43.6 \times(0.3)^2} \mathrm{rad} / \mathrm{s}^2$\\\\=4.4342 \mathrm{rad} / \mathrm{s}^2$[/tex]

[tex]$\quad \Delta \theta=0.320 \mathrm{rev}=0.320 \times 215 \mathrm{rad}$[/tex]

An Angular velocity at that instant [tex]$=\omega_f > $[/tex]

[tex]$$\begin{aligned}& \omega_i=0 \\& \text { using } \omega_f^2=\omega_i^2+2 \alpha \Delta \theta \\& \omega_f^2=0+2 \times 4.4342 \times 0.320 \times 2 \pi \\& \omega_f=\sqrt{2 \times 4.4342 \times 0.320 \times 2\Pi} \mathrm{rad} / \mathrm{s} \\& \omega_f=4.22 \mathrm{rad} / \mathrm{s}\end{aligned}$$[/tex]

The tangential velocity of point on [tex]$n m=\omega_f r$[/tex]

[tex]$$\begin{aligned}& V=4.22 \times 0.3 \mathrm{~m} \\& V=1.266 \mathrm{~m} / \mathrm{s}\end{aligned}[/tex]

The complete question should be:

A uniform disk with mass 43.6 kg and radius 0.300 m is pivoted at its centre about a horizontal, frictionless axle that is stationary. the disk is initially at rest, and then a constant force 29.0 n is applied tangent to the rim of the disk. What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.320 revolutions?

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If you traveled for 25hours with an average speed of 48miles/hours, the distance travel

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If you traveled for 25 hours with an average speed of 48 miles/hour, the distance traveled is 25 * 48 = 1200 miles. This can be found by multiplying the number of hours traveled by the average speed to get the total distance traveled.

A ball rolls down a curved ramp as shown in the diagram below. Which dotted line best represents the path of the ball after leaving the ramp? A) A B) B C) C D) D

Answers

The movement of the ball down the ramp can be shown by option B.

What is the path of the ball?

We know that the question has clearly said that the ramp was curved and this is going to help in shaping our discussion. In this case, it is clear that the the ball  rolls down a curved ramp as shown. The movement of the ball can be typified by the use of dotted lines.

In this case, we can see that the ball is going to continue to move in a curved path as the ball is moving down the ramp as has been shown in the image.

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