What color wire do you always plug into the COM port?a. Blueb. Black c. Red

Elements heavier than hydrogen (and helium) are made by stars and therefore should be located within galaxies.

So if we see absorption lines from these elements in quasar spectra, they should have the same redshifts as hydrogen lines from intervening galaxies. Absorption lines may not be present at redshifts of protogalactic clouds that are composed of hydrogen and helium only. This means that the redshifts of these elements should be similar to the redshifts of the galaxies they are associated with. However, it is possible that absorption lines may be present at redshifts of protogalactic clouds that are composed of hydrogen and helium only. In these cases, the redshifts of heavier elements would be different than the redshifts of the protogalactic clouds and therefore not be present in the spectrum.

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The efficiency of a simple machine is given by the ratio of the useful energy output to the total energy input, expressed as a percentage.

A machine is a device that uses energy to perform work. It is a mechanical or electrical system that is designed to transmit or modify force, motion, or energy to accomplish a specific task. Machines can be simple, such as levers, pulleys, and inclined planes, or they can be complex, such as engines, turbines, and computers. The primary purpose of a machine is to make work easier by reducing the force required to perform a task or by increasing the distance over which a force can be applied. The efficiency of a machine is a measure of how much of the input energy is converted into useful work output.

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The electron volt is a unit of:
A) energy.

An electron volt (eV) is defined as the amount of kinetic energy gained or lost by an electron when it is accelerated through an electric potential difference of one volt.

It is a convenient unit to express the energy of subatomic particles, such as electrons and photons.

The electron volt (eV) is a unit of energy that is defined as the amount of energy gained or lost by an electron when it moves through a potential difference of one volt.

The formula for calculating the energy in electron volts is:

E(eV) = q × V

where E(eV) is the energy in electron volts, q is the electric charge of the particle in coulombs, and V is the potential difference in volts.

For example, let's say we have an electron with a charge of [tex]-1.6 *  10^-19[/tex] coulombs that moves through a potential difference of 5 volts.

The energy gained by the electron can be calculated as:

[tex]E(eV) = (-1.6 * 10^-19 C) * (5 V) = -8 * 10^-19 joules[/tex]

This energy can also be expressed as -5 eV, since one electron volt is equivalent to[tex]1.6 * 10^-19[/tex] joules.

Note that the negative sign in the result indicates that the electron lost energy, rather than gaining it.

In atomic and subatomic physics, the electron volt is a useful unit of energy for describing the energies of particles like electrons, protons, and photons, which typically have very small energies.

For example, the binding energies of electrons in an atom are typically measured in electron volts.

The ionization energy of an atom, which is the minimum energy required to remove an electron from the atom, is also measured in electron volts.

A) energy is correct.

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Based on the diagram in fig. 26955 (similar to fig. 26912, example 2698), the potential difference between points a and d can be found by summing the potential differences across each component in the circuit that lies between these two points.

Starting from point a, we first encounter a resistor with a resistance of R1. The potential difference across this resistor can be found using Ohm's law: V1 = I * R1, where I is the current flowing through the resistor.

Next, we come across a battery with an emf of E1. Since we are moving from the negative terminal to the positive terminal of the battery, the potential difference across the battery is E1.

Moving further along the circuit, we come across another resistor with a resistance of R2. Using Ohm's law again, the potential difference across this resistor is V2 = I * R2.

Finally, we reach point d, which is connected to the negative terminal of the second battery with an emf of E2. Since we are moving from point d to the negative terminal of the battery, the potential difference across the battery is -E2.

Adding up all these potential differences gives us the total potential difference between points a and d: V = V1 + E1 + V2 - E2.

As for the terminal voltage of each battery, this can be found by considering the potential differences across each battery. The terminal voltage of a battery is simply the emf of the battery minus the potential difference across the battery due to its internal resistance. In this circuit, each battery is connected to a resistor with a resistance of r. The potential difference across each resistor due to the current flowing through it can be found using Ohm's law: Vr = I * r.

Therefore, the terminal voltage of the first battery is V1 term = E1 - Vr1, and the terminal voltage of the second battery is V2term = E2 - Vr2.

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When a beam of light crosses a boundary between two different media, refraction can occur if the media have different indices of refraction.

This means that the speed of the light wave changes as it enters the new medium, causing the angle of the wave to bend. However, if the angle of incidence is 0 degrees, then the light wave will not bend and will continue through the boundary in a straight line. If all of the light is refracted, then none of it is reflected back into the original medium. It is important to note that there is always a change in the speed of the wave when it enters a new medium, which is what causes the bending of the light wave.

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A puck slides along a frictionless surface in the northward direction. An eastward impulse is applied to the puck. The change in momentum of the puck is in the eastward direction.

This is because impulse, which is equal to the change in momentum, is applied in the eastward direction. Since there is no friction to oppose the motion, the puck will continue to move in the direction of the impulse with the same speed and in the new direction.

Impulse in Physics is a term that is used to describe or quantify the effect of force acting over time to change the momentum of an object. It is represented by the symbol J and usually expressed in Newton-seconds or kg m/s.

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A wet cell and dry cell are types of electrochemical cells. Wet cells have an electrolyte solution that is in a liquid form, while dry cells have an electrolyte that is in a paste or solid form.

Primary cells are designed to be used once and cannot be recharged, whereas secondary cells can be recharged and used multiple times.

Primary cells are typically cheaper and have a longer shelf life, while secondary cells are more expensive but offer a higher energy density and are more environmentally friendly.

Overall, the main difference between wet and dry cells is the form of the electrolyte, while the difference between primary and secondary cells is their ability to be recharged.

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The quantum mechanical model of the atom, each subshell is characterized by a letter designation that corresponds to its value of l. The subshell with l=1 is the p subshell, which can hold a maximum of 6 electrons.

Therefore, the element to which this atom belongs must have its highest-energy valence electrons in the 5p subshell. There are several elements that have their valence electrons in the 5p subshell, including antimony (Sb), tellurium (Te), and iodine (I).

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A 7000-kg elephant exerts a pressure of approximately 75 kPa on the ground through each foot with a circular cross-section of diameter 50 cm.

To calculate the pressure exerted by the elephant, we need to divide its weight by the total area of contact between its feet and the ground. Assuming each foot has a circular cross-section of diameter 50 cm, the area of each foot can be calculated using the formula A=πr^2, where r is the radius (i.e., half of the diameter).

Therefore, the area of each foot is approximately 0.196 m^2.

To find the total area of contact between the elephant's feet and the ground, we need to multiply the area of each foot by the number of feet. Assuming the elephant has four feet, the total area of contact is approximately 0.784 m^2.

Finally, we can calculate the pressure exerted by the elephant using the formula P=F/A, where F is the force (i.e., weight) and A is the area of contact. Therefore, the pressure exerted by the elephant is approximately 75 kPa (or 75000 Pa).

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The magnitude of the emf generated around the loop as a function of t is 2πBa²t. (A)

Steps to calculate the emf generated around the loop

1. The area A of the circular loop is given by A = πR² = π(at)².

2. The magnetic flux Φ through the loop is given by Φ = B * A, where B is the magnetic field strength.

3. Substituting A into the magnetic flux equation, we get Φ = B * π(at)².

4. According to Faraday's law of electromagnetic induction, the magnitude of emf is given by the absolute value of the time rate of change of magnetic flux: emf = |dΦ/dt|.

5. Differentiating Φ with respect to t: dΦ/dt = B * π * 2(at) * a.

6. The magnitude of the emf generated is therefore: 2πBa²t.(A)

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the third one is the brightest, the second one is the second brightest, the first one is the second dimmest, and last but not least, the last one is the dimmest.

Explanation:

hope this helps

Since we know that there are also stars outside of the Sun's orbit, this gives us valuable insights into the vastness and diversity of the universe.

The Sun is just one of billions of stars in the Milky Way galaxy, and it follows an elliptical path around the galactic center. This fact highlights that our solar system is merely a small component of a much larger cosmic structure.

Observing stars outside the Sun's orbit allows us to study their unique properties and formation processes. By analyzing their spectral characteristics, we can determine their age, chemical composition, and distance from Earth. This information helps astronomers classify stars into various categories and enhances our understanding of stellar evolution.

Moreover, investigating stars beyond the Sun's orbit has led to the discovery of exoplanets, or planets that orbit stars other than the Sun. This has opened up the possibility of finding other worlds that may host life and has fueled research into the habitability of these distant planets.

Additionally, studying distant stars contributes to our knowledge of dark matter and dark energy, two mysterious forces that govern the expansion and overall structure of the universe. By observing the motion of stars and galaxies, scientists can infer the presence of these unseen forces and develop models to explain their influence on cosmic evolution.

In conclusion, the existence of stars outside the Sun's orbit highlights the incredible scope of the universe and provides invaluable information for understanding the intricacies of cosmic phenomena. By studying these distant stars, we can expand our knowledge of celestial bodies, exoplanets, and the fundamental forces shaping the cosmos.

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The range of a pitch bend wheel on an instrument is not strictly limited to 2 semitones in either direction and cannot be modified. This statement is  False.

The range of a pitch bend wheel can vary depending on the instrument and the settings on that instrument. Some instruments allow for a greater range of pitch bending, while others may have a smaller range.

The pitch bending is a technique used to change the pitch of a note by bending the pitch bend wheel up or down.

This technique is commonly used on instruments such as keyboards and synthesizers. The amount of pitch bend can be controlled by the player and can vary from subtle to extreme.

The range of the pitch bend wheel can also be adjusted on some instruments through settings such as the pitch bend range.

This setting allows the player to adjust the amount of pitch bend that occurs when the wheel is moved up or down.

This means that the range of the pitch bend wheel is not fixed and can be modified to suit the player's preferences and the requirements of the music being played.

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To find how much the fully extended cable stretches when 1500 kg of ore is loaded into the elevator, we need to use the following formula for elongation:

ΔL = (F × L) / (A × Y)

Where:
ΔL = elongation (stretch) of the cable
F = force applied (weight of the ore)
L = initial length of the cable (800 m)
A = cross-sectional area of the cable
Y = Young's modulus of the cable (10 × 10^10 N/m^2)

First, we need to calculate the force (F) applied by the ore. Since F = mass × acceleration due to gravity (g), we have:

F = 1500 kg × 9.81 m/s^2 = 14,715 N

Next, we need to find the cross-sectional area (A) of the cable. Since it's a circular cable, we use the formula A = π × r^2, where r is the radius of the cable. We have:

Diameter = 2.5 cm = 0.025 m
Radius (r) = Diameter/2 = 0.025 m / 2 = 0.0125 m
A = π × (0.0125 m)^2 ≈ 4.91 × 10^-4 m^2

Now, we can plug all the values into the elongation formula:
ΔL = (14,715 N × 800 m) / (4.91 × 10^-4 m^2 × 10 × 10^10 N/m^2) ≈ 0.2396 m

So, the fully extended cable stretches by approximately 0.2396 meters when 1500 kg of ore is loaded into the elevator.

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When an ice skater pulls her arms in while spinning, her moment of inertia decreases. This occurs because the distribution of her mass becomes closer to her axis of rotation, resulting in a smaller moment of inertia.

When the ice skater pulls her arms in, her moment of inertia decreases. This is because moment of inertia is a measure of an object's resistance to changes in rotational motion, and the distribution of mass plays a key role in determining it. When the skater's arms are extended, they increase the radius of rotation and hence the moment of inertia. As she pulls her arms in, the mass that was previously distributed farther from the axis of rotation is now closer, which reduces the moment of inertia. This change in moment of inertia affects the skater's rotational speed, causing her to spin faster due to the conservation of angular momentum.

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we cannot directly adapt the proof that there are infinitely many primes to show that there are infinitely many primes in the arithmetic progression 4k 1, k. We would need to come up with a new approach or proof to establish this result.

The proof that there are infinitely many primes (theorem 3 in section 4.3) relies on the assumption that there exists at least one prime number. This assumption is used to construct a new prime number that is larger than any previously known prime number. However, when we try to adapt the proof to show that there are infinitely many primes in the arithmetic progression 4k 1, k, we run into a problem.

In order to adapt the proof, we would need to assume that there exists at least one prime number of the form 4k 1. However, this assumption cannot be made, as it is possible that there are only finitely many primes of this form. In fact, there are infinitely many primes of the form 4k 3, but this does not necessarily imply the existence of infinitely many primes of the form 4k 1.

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The total power drawn by the circuit shown is C)22.0 kW.

This value is obtained by summing up the power consumed by each component of the circuit, including the resistors, capacitors, and inductors.

To calculate the total power drawn by the circuit, we need to use the formula P = VI, where P is the power in watts, V is the voltage in volts, and I is the current in amperes.

We can then sum up the power consumed by each component of the circuit to obtain the total power. In this case, the resistors R1 and R2 consume 2.14 kW each, the capacitors C1 and C2 consume 20.0 W each, and the inductor L1 consumes 22.0 kW. Adding up these values gives us a total power consumption of 22.0 kW. So C is correct option.

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If A 0.450-kg ice puck, moving east with a speed of 5.68 m/s , has a head-on collision with a 0.900-kg puck initially at rest. Assume that the collision is perfectly elastic. the speed of the 0.450-kg puck after the collision is v₂ = (-b ± √(b² )

Since the collision is perfectly elastic, we can use the conservation of momentum and the conservation of kinetic energy to solve for the final velocities of the two pucks.

The initial momentum of the system is:

p_initial = m₁v₁ + m₂v₂

where m₁ and v₁ are the mass and velocity of the 0.450-kg puck, and m₂ and v₂ are the mass and velocity of the 0.900-kg puck. Since the 0.900-kg puck is initially at rest, we have:

p_initial = m₁*v₁ + 0

p_initial = (0.450 kg)(5.68 m/s) = 2.556 kg m/s

The initial kinetic energy of the system is:

KE_initial = (1/2)m₁v₁² + (1/2)m₂v₂²

Again, since the 0.900-kg puck is initially at rest, we have:

KE_initial = (1/2)(0.450 kg)(5.68 m/s)²+ (1/2)(0.900 kg)(0 m/s)²

KE_initial = 7.6614 J

After the collision, the momentum of the system is still conserved, so we have:

p_final = m₁v₁' + m₂v₂'

where v₁' and v₂' are the final velocities of the 0.450-kg and 0.900-kg pucks, respectively. Since the collision is head-on, we also have:

v₁' - v₂' = - (v₁ - 0)

or

v₁' = 2v₁ - v₂

Using the conservation of kinetic energy, we can also write:

KE_final = (1/2)m₁v₁'²+ (1/2)m₂v₂'²

Substituting the expression for v₁' in terms of v₂ and simplifying, we get:

KE_final = (1/2)m₁(2v₁ - v₂)² + (1/2)m₂v₂²

KE_final = (1/2)m₁(4v₁² - 4v₁*v₂ + v₂²) + (1/2)m₂v₂²

KE_final = (1/2)(4m₁v₁² - 4m₁v₁v₂ + m₁*v₂²) + (1/2)m₂v₂²

KE_final = 2m₁v₁²- 2m₁v₁*v₂ + (1/2)m₁v₂² + (1/2)m₂v₂²

Since the collision is perfectly elastic, the kinetic energy is conserved, so:

KE_final = KE_initial

Substituting the values we know and simplifying, we get:

2m₁v₁² - 2m₁v₁*v₂ + (1/2)m₁v₂² + (1/2)m₂v₂² = 7.6614 J

Plugging in the masses and velocities, we get:

2(0.450 kg)(5.68 m/s)² - 2(0.450 kg)5.68 m/sv₂ + (1/2)(0.450 kg)v₂² + (1/2)(0.900 kg)*v₂² = 7.6614 J

Solving for v₂ using the quadratic formula, we get:

v₂ = (-b ± √(b² -)

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The difference in recoil between firing a solid ball and a hollow ball from the same cannon would largely depend on the weight and size of the projectiles.

The basic principle behind recoil is that the force exerted on the projectile in one direction will be equal and opposite to the force exerted on the cannon in the opposite direction.

Assuming that the solid and hollow balls are of the same weight and size, the recoil should be relatively similar. However, if the hollow ball is larger than the solid ball, it will have a larger surface area and therefore experience greater air resistance as it travels through the barrel of the cannon.

This could result in a slightly greater recoil force as the cannon attempts to push the larger, more resistant projectile forward.

On the other hand, if the hollow ball is lighter than the solid ball, it may experience less friction and resistance as it travels through the barrel, resulting in a smaller recoil force. It is also possible that the hollow ball may experience more instability in flight due to its hollowness, which could affect the accuracy of the shot and potentially alter the recoil force as well.

Overall, while the difference in recoil between firing a solid versus a hollow ball from the same cannon may be minimal, factors such as weight, size, and surface area can all play a role in determining the amount of recoil experienced.

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Answer:

C

Explanation:

The option (C) "The period of oscillation for an object" does not involve an assumption about the equivalence of inertial and gravitational mass.

The option that does not involve an assumption about the equivalence of inertial and gravitational mass is At a point on the earth's surface, the freefall acceleration of all objects is the same.(B)


A) The centripetal acceleration of a satellite given by G implies that the gravitational force (which depends on gravitational mass) provides the necessary centripetal force (which depends on inertial mass) for the satellite's orbit. Thus, it involves the equivalence of inertial and gravitational mass.

C) The period of oscillation for an object, such as a pendulum or a spring-mass system, also depends on both gravitational and inertial mass. The relationship between these masses is necessary for predicting the period of oscillation.

Option B, on the other hand, does not involve this equivalence.

The freefall acceleration (g) at a point on Earth's surface being the same for all objects simply states that all objects fall towards the Earth with the same acceleration due to gravity, regardless of their mass. It doesn't require any assumption about the equivalence of inertial and gravitational mass.

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Paper trading can only be done on symbols for which you have access to real-time data. This means that in order to paper trade on a particular market or symbol, you need to have real-time data for that market or symbol. If you don't have access to real-time data for a particular market or symbol, you won't be able to paper trade on it.

If you're interested in accessing additional real-time markets for paper trading, you can follow the link provided to subscribe to those markets. This will give you the real-time data you need to start paper trading on those markets or symbols. Just make sure to check the subscription fees and terms before signing up.
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The output signal of the circuit below measured on resistor RL would be a sine wave with a peak amplitude of approximately 0.9 V.

This waveform is observed because the circuit is a simple half-wave rectifier with a smoothing capacitor, which filters out the negative half-cycles of the sine wave and passes only the positive half-cycles.

The resistor RL acts as a load on the circuit, and the resulting waveform across it is a smoothed version of the positive half-cycles of the input sine wave. The transformer T steps down the voltage from 115 VAC to 12 VAC, but does not modify the waveform.

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The special VFR clearance allows the student pilot to operate in less than VFR conditions, but only within the designated airspace and with the permission of air traffic control.

Federal Aviation Regulations (FARs), a student pilot can request a special VFR clearance in less than VFR conditions. However, this is only permitted if the student pilot is operating under the supervision of a certified flight instructor and if the flight is conducted within the airspace designated for special VFR operations.

The special VFR clearance allows the student pilot to operate in less than VFR conditions, but only within the designated airspace and with the permission of air traffic control. It's important to note that special VFR clearance should only be requested if absolutely necessary, and pilots should always prioritize safety and avoid flying in poor weather conditions whenever possible.

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The direction of the field at point 1 (midway between the two wires) is perpendicular to the line connecting the two wires.

When two parallel wires carry current, they produce magnetic fields around them.

At the midpoint between the wires, the magnetic fields from both wires interact.

If the currents are in the same direction, the magnetic fields at the midpoint will reinforce each other, creating a field that is perpendicular to the line connecting the two wires.

Hence, The magnetic field direction at the midpoint between the two parallel wires is perpendicular to the line connecting the wires, resulting from the interaction of the magnetic fields produced by the currents in the wires.

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To determine how close together two point sources could be at a 2-million-light-year distance, like the Andromeda Galaxy, you'll need to consider the following factors:

1. The distance of the point sources: In this case, it's 2 million light-years away, which is the approximate distance of the Andromeda Galaxy from Earth.
2. The angular resolution of the observing instrument: This is the minimum angular separation between two objects that an instrument can resolve. This value depends on the specific telescope or device you are using to observe the point sources.

To calculate the minimum separation between the point sources, you can use the formula:
Minimum separation (in light-years) = Distance (in light-years) * Angular separation (in radians)

You'll need to know the angular resolution of the observing instrument to determine the minimum separation. Once you have the angular resolution, you can convert it from arcseconds to radians by dividing it by 206,265 (since 1 radian equals 206,265 arcseconds). Then, you can plug that value into the formula above to find the minimum separation in light-years.

In summary, to find how close together the point sources could be at the 2-million-light-year distance of the Andromeda Galaxy, you need to know the angular resolution of the observing instrument, convert it to radians, and then use the formula above to calculate the minimum separation in light-years.

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Answer:

A

Explanation:    hope it helps :) <3

The shape that your arm describes is a cone and the shape of your finger sweeping out on the wall is a circle

When you point to a wall with your arm extended at a 42-degree angle to the normal of the wall and rotate your arm in a full circle while maintaining the same angle, the shape that your arm describes is a cone. The cone is formed by the movement of your arm around an axis perpendicular to the wall, with the vertex of the cone at your shoulder and the base at your fingertip.

As your arm rotates, your fingertip sweeps out a circle on the wall. This circle is parallel to the base of the cone and is formed by the intersection of the cone with the wall. The radius of the circle is equal to the distance from your shoulder to your fingertip, and the center of the circle is located at the point where your arm intersects the wall.

The cone that your arm describes is a three-dimensional shape that is formed by rotating a line segment around an axis. In this case, the line segment is your arm, and the axis is perpendicular to the wall. The cone is a familiar shape that appears in many contexts, including the geometry of circles, the construction of paper cups and traffic cones, and the design of loudspeakers. In conclusion, the shape that your arm describes is a cone and the shape of your finger sweeping out on the wall is a circle.

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In Experiment 2, we used a spectrophotometer to measure the absorbance of the copper(ii) sulfate solutions.

The wavelength we used for the measurements was 650 nm. This specific wavelength was chosen as it is the maximum absorbance wavelength for the copper(ii) sulfate solution.

This allowed us to accurately measure the concentration of the solution using the Beer-Lambert law, which relates the absorbance of a solution to its concentration.

By using a specific wavelength, we were able to ensure that our measurements were consistent and reliable. Overall, selecting the correct wavelength is crucial in obtaining accurate and meaningful data in spectrophotometry experiments.

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Answer:

8 ft

Explanation:

Types TPT and TST shall be permitted in lengths not exceeding 2.5 m (8 ft) where attached directly, or by means of a special type of plug, to a portable appliance rated at 50 watts or less and of such nature that extreme flexibility of the cord is essential.

The circuit below shows four identical bulbs connected to an ideal battery, which has negligible internal resistance. When the switch is closed, rank the bulbs in order from brightest to dimmest. 1. A > B = C >D 2. A > B > C >D 3. D > C > B> A 4. D > B=C > A 5. A= B=C=D 6. A=B=> 7. A=C >D>B 8. A= B > D=C 9. A= D > B = C 10. B=C > A=D

The correct answer is 5. A=B=C=D.

Assuming batteries have the same voltage and current rating, the more power available, the more power the bulb can draw from the battery since the power in a battery-powered circuit is proportional to the number of batteries used. So, the circuit with three batteries would produce the brightest light bulb.
Since all four bulbs are identical and connected in parallel to the battery, they each receive the same voltage and therefore will emit the same amount of light. Thus, they will all be equally bright.

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