If the magnitude of the drift velocity of free electrons in a copper wire is 7.12 10-4 m/s, what is the electric field in the conductor?
___?V/m

Answer:

c

Explanation:

force is how hard it is pulled or pushed

The current provided in each branch for 12V circuit are 6 A and 4 A.

The total current provided for the 12 V circuit is 10 A.

The current provided in each branch for 9V circuit are 4.5 A, 3 A and 9 A.

The total current provided for the 12 V circuit is 16.5 A.

The current provided in each branch of the parallel circuit is calculated as follows;

For the 12V circuit;

I = V/R

where;

branch 1 = 12/2 = 6 A

branch 2 = 12/3 = 4 A

Total resistance;

1/Rt = 1/2 + 1/3

1/Rt = 5/6

Rt = 6/5 = 1.2 ohm

Total current is calculated as;

I_t = 12 V/1.2 ohm = 10 A

For the 9 V circuit:

branch 1 = 9/2 = 4.5 A

branch 2 = 9/3 = 3 A

branch 3 = 9/1 = 9 A

Total resistance;

1/Rt = 1/2 + 1/3 + 1/1

1/Rt = 11/6

Rt = 6/11 = 0.545 ohms

Total current is calculated as;

I_t = 9 V/0.545 ohm = 16.5 A

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Answer:

\(\huge\boxed{\sf F = 14.52\ N}\)

Explanation:

Mass = m = 3,630 g = 3.63 kg

Acceleration = a = 4 m/s²

Force = F = ?

F = ma

Put the given data in the above formula.

F = (3.63)(4)

\(\rule[225]{225}{2}\)

Explanation:

Hope its helps u

if wrong then sry

The time required for the rock to reach the ground is 1.11 second.

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

Initial height of the rock: h = 6.00 m.

Acceleration due to gravity: g = 9.8 m/s²

Let,  the time required for the rock to reach the ground is = t.

Then:

h = 1/2 × gt²

t = √(2h/g)

= √ {(2 × 6.00)/9.8} second.

= 1.11 second.

Hence, the time required for the rock to reach the ground is 1.11 second.

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Answer:

Waves move through different vmaterials

A:  Waves travel at different speeds through different materials.

Answer:

Ascaris and Arthropods are both types of organisms found in the animal kingdom. Ascaris are parasitic worms, commonly referred to as roundworms, which can be found in warm climates all over the world. Arthropods, on the other hand, are a large group of animals, including insects, arachnids, and crustaceans, that typically have jointed legs and a hard exoskeleton. Arthropods are found in almost all environments, from oceans to deserts to the tops of mountains.

Explanation:

Yes, under certain conditions, it is possible for small metal objects to float on water due to surface tension. This phenomenon is known as "floatation on surface tension" or "floating on a water surface."

When an object is placed on the surface of water, it experiences several forces acting on it, including gravity and surface tension. If the weight of the object is balanced by the upward force exerted by surface tension, the object can float on the water surface.

The ability of an object to float on water depends on its density compared to the density of water. If the object is less dense than water, it will float; if it is more dense, it will sink.

In the case of a small diameter metallic rod, the density of the material is given as 7801 kg/m³. This density is significantly higher than the density of water, which is approximately 1000 kg/m³. Therefore, a solid metallic rod made of this material would sink in water rather than float.

To achieve floatation on surface tension, you would typically require objects with lower densities, such as small insects or certain types of lightweight materials that can distribute their weight over a larger surface area, allowing surface tension to support them.

It's important to note that while surface tension can support small objects under specific conditions, it has limitations. Once the size or weight of the object exceeds certain thresholds, the surface tension forces become insufficient to keep it afloat, and it will sink.

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Answer:

\(we \: know \: energy \: = \: force \: \times distance \\ e = f \times d \\ so \: f \: = \frac{e}{d} \\ so \: th \: force \: here \: = (\frac{2670}{3}) newton \\ = 890newton\)

Hope it helps

If a projectile travels through air, it loses some of its kinetic energy due to air resistance. Some of this lost energy is converted into heat and sound as the projectile interacts with the air molecules.

If a projectile travels through air, it loses some of its kinetic energy due to air resistance. This lost energy is primarily converted into heat and sound as the projectile interacts with the air molecules. The air resistance creates a drag force that acts opposite to the direction of the projectile's motion. As the projectile moves through the air, the drag force opposes its velocity, causing a deceleration and reducing its kinetic energy. This energy is dissipated in the form of heat due to the friction between the projectile and the surrounding air. Additionally, the disturbance caused by the projectile moving through the air generates sound waves, resulting in the conversion of some kinetic energy into sound energy. Overall, the kinetic energy lost to air resistance manifests as heat and sound during the projectile's flight.

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Given :- A resistor of 150 ohm, hence Resistance (R) = 150 ohm

Potential Difference (v) = 24 V

Current (I) = ?

V = IR

24 = I × 150

I = 24/150

I = 0.16 ampere

hope it helps!

Answer:

t = 1.33 seconds

Explanation:

Given that,

The mass of the ball, m = 5 kg

Initial speed, u = 8 m/s

Final speed, v = 12 m/s

Force, F = 15 N

We need to find the time for long the force is applied. Let the time be t. The force is given by :

\(F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{5\times (12-8)}{15}\\\\=1.33\ s\)

So, the force is applied for 1.33 seconds.

Given, acceleration of a particle on the x-axis is a = -v². The initial position and velocity are x(0) = 1, x'(0) = 2. We have to find the position, velocity, and acceleration as functions of time x(t), v(t), a(t). We can use the formulas below to solve the given problem. Position as a function of time, x(t)To find the position as a function of time x(t), we need to integrate the given acceleration function with respect to time.taking negative sign common we get,a = -v²dv/dx = -v dv = -v dx dv/v² = -dx x = ∫ dx/x² + cputting limits and solving the integral.

we get,x = 1/(v + c₁)Taking derivative both side w.r.t t,x'(t) = (-1/(v+c₁)²) . dv/dtTo find the value of constant c₁, we use the initial velocity,x'(0) = 2 = (-1/(v(0)+c₁)²) . dv/dt(2/1) = dv/dt ..... (1)Integrating both sides of the above equation with respect to t, we getv(t) = -2t + Cwhere C is the constant of integration. Substituting the value of v(t) in the equation x = 1/(v + c₁), we getx(t) = 1/(-2t + C + c₁)Velocity as a function of time, v(t).

The velocity as a function of time, v(t) is given byv(t) = -2t + C, where C is the constant of integration.Initial velocity is v(0) = -2(0) + C = C = 2Substituting this value in the above equation, we getv(t) = -2t + 2Acceleration as a function of time, a(t)Acceleration as a function of time is given bya(t) = -v²(t) = -(-2t+2)²= -(4t² - 8t + 4)= -4(t² - 2t + 1) = -4(t-

1)²Therefore, Position as a function of time, x(t) is x(t) = 1/(-2t + C + c₁).Velocity as a function of time, v(t) is v(t) = -2t + 2.Acceleration as a function of time, a(t) is a(t) = -4(t-1)².

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Answer: D

Explanation:

If Rhea is inspired and not too focused on comparing abilities, Biles is a good role model.

Quantum tunneling can be described as a quantum mechanical process where wavefunctions can penetrate through a potential barrier.

The transmission via the potential barrier can be finite and also depend upon the barrier width as well as barrier height. The wave functions have the probability of disappearing on one side and reappear on the remaining side.

The 1st derivative of the wave functions is generally continuous. In the steady state, the probability flux is spatially uniform and no wave or particle is eliminated. Tunneling can occur with barriers of thickness about 1 to 3 nm and smaller.

Quantum tunneling plays a crucial role in phenomena such as nuclear fusion and α-radioactive decay of atomic nuclei. Quantum tunneling has applications in the tunnel diode, in scanning tunneling microscope, and in quantum computing.

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Quantum tunneling is a phenomenon in quantum mechanics where wavefunctions can pass through a potential barrier. It is a process that defies classical physics, where particles would be expected to bounce back when they encounter a barrier.

In quantum mechanics, particles are described by wavefunctions, which are mathematical representations of their quantum states. These wavefunctions can extend into regions that are classically forbidden, such as inside a potential barrier.

The probability of a particle "tunneling" through the barrier depends on various factors, including the width and height of the barrier. The wavefunctions have a certain probability of disappearing on one side of the barrier and reappearing on the other side.

One important characteristic of quantum tunneling is that the first derivative of the wavefunctions is generally continuous. This means that there is a smooth transition between the regions before and after the barrier. In the steady state, the probability flux is spatially uniform, meaning that no waves or particles are lost during the tunneling process.

Quantum tunneling has significant implications in various areas of physics. It plays a crucial role in phenomena such as nuclear fusion and alpha-radioactive decay of atomic nuclei. It also has practical applications, such as in tunnel diodes, scanning tunneling microscopes, and quantum computing.

Quantum tunneling is a fascinating phenomenon in quantum mechanics where particles can pass through potential barriers that classical physics would consider impossible to overcome.

In classical physics, particles are confined to their energy levels and cannot pass through energy barriers unless they have enough energy to overcome them. However, in quantum mechanics, particles can exhibit wave-like behavior and their wavefunctions can extend beyond the physical boundaries we would expect them to be confined to.

When a particle encounters a potential barrier, there is a probability that its wavefunction can penetrate through the barrier and appear on the other side. This means that even if the particle does not have enough energy to overcome the barrier classically, there is still a chance that it can "tunnel" through the barrier and continue its motion on the other side.

The probability of tunneling depends on factors such as the width and height of the potential barrier. In some cases, the wavefunction can completely disappear on one side of the barrier and reappear on the other side.

Quantum tunneling is not limited to any specific system but can occur in a wide range of phenomena. It has been observed in nuclear fusion, where particles overcome the Coulomb repulsion barrier and fuse together. It also plays a role in the alpha decay of atomic nuclei. Additionally, quantum tunneling has practical applications in devices like tunnel diodes, which exploit this phenomenon to create unique electrical behavior.

Overall, quantum tunneling is a fundamental concept in quantum mechanics that allows particles to defy classical limitations and pass through barriers that would be impossible to overcome classically.

Answer:

1.0x10^7 Pa or 1,460 psi.

Explanation:

2000N is applied to an area of (0.02m)*(0.01m).  Pressure is in units of N/m^2.  1 N/m^2 is 1 Pascal (Pa).

 (2000N)/(0.002 m^2) = 1.0x10^7 Pa

This is equal to 1,460 psi

Answer:

1. Flexibility.

2. Muscular Fitness

3. Cardiovascular Fitness

4. Cardiovascular Fitness

5. Flexibility

6. Muscular Fitness

7. Cardiovascular Fitness

Answer:

v = -5 m/s

Explanation:

It is given that,

The initial velocity of the ball, u = 25 m/s

We need to find its velocity after 2 seconds in the air.

Let v is the velocity of the ball after 2 seconds. Using equation of motion to find it.

v = u + at

Here, a = -g

v = u -gt

Putting all the values,

v = 25 - (10)(3)

= 25-30

= -5 m/s

So, the velocity of the ball after 3 seconds is 5 m/s and it is in downward direction.

Answer:130.7s

Explanation:

The charge of the particle in this situation is positive. When a charged particle is immersed in a uniform magnetic field, it experiences a force known as the Lorentz force, which is given by the equation:

F = q(v × B)

Where F is the Lorentz force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field.

In this case, the particle is moving along a circular path in the counter-clockwise direction, which means that the force acting on the particle is directed towards the center of the circle. This force is known as the centripetal force, and it is given by the equation:

F = mv^2/r

Where m is the mass of the particle, v is the velocity of the particle, and r is the radius of the circle.

Since the Lorentz force and the centripetal force are equal in magnitude and opposite in direction, we can equate the two equations to get:

q(v × B) = mv^2/r

Rearranging the equation and solving for q gives:

q = mv^2/(rB)

Since the particle is moving in the counter-clockwise direction, the velocity vector v is directed tangentially to the circle, and the magnetic field vector B is directed out of the page. The cross product of these two vectors is directed towards the center of the circle, which means that the charge of the particle must be positive in order for the Lorentz force to be directed towards the center of the circle.

Therefore, the charge of the particle in this situation is positive.

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Answer: The energy to move the fluid is provided by the pressure on the fluid surface.

Explanation:  The frictional losses in the suction pipework and rises in the suction pipework system will reduce the fluid pressure at the pump inlet. If the pump inlet connection is removed the fluid will not flow out of the suction pipework. Hope this helps! :)

Les molécules sont arrangées autrement, ce qui modifie l'aspect ou l'état de l'espèce chimique. Un mélange est une substance composée de plusieurs espèces chimiques qui ne réagissent pas ensemble, au contraire de la transformation chimique. Il s'agit donc d'une transformation physique, qui ne change pas la matière.

The most popular electrode diameters range from small sizes, typically around 1 mm, to larger sizes, often up to 10 mm or more.

Electrodes are commonly used in various fields such as medical applications, electrical engineering, and material science. The diameter of an electrode plays a crucial role in its functionality and application. Smaller electrode diameters, typically around 1 mm, are commonly used in precision applications where fine control and high resolution are required. These smaller electrodes are suitable for intricate tasks such as microsurgery, microelectronics, and microfabrication.

On the other hand, larger electrode diameters are utilized for applications that require higher current-carrying capacity and greater durability. These larger electrodes, which can range from several millimeters to centimeters in diameter, are commonly used in industries such as welding, metal fabrication, and power generation. The larger diameter allows for efficient heat dissipation, better electrical conductivity, and improved mechanical stability.

It's important to note that the specific range of popular electrode diameters can vary depending on the specific application, industry standards, and technological advancements. Therefore, it is always recommended to consult the relevant guidelines or industry practices to determine the appropriate electrode diameter for a particular use case.

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Answer:

Negative initial velocity doesn't equal the final velocity of the ball at the maximum range of the ball.

The final velocity of the ball at the maximum range is equal to zero, since at that point the ball has reached its highest point and is no longer moving upward, it is now moving downward.

In physics, when an object is launched upward, its initial velocity is positive and it slows down as it reaches the highest point, then its velocity becomes negative as it starts to fall. The final velocity of the object is zero at the highest point as it is no longer moving upward and it's not yet moving downward.

In the case of a ball being thrown, it's initial velocity is positive when it's thrown, and it slows down as it reaches the highest point, then it's final velocity is zero at the highest point and negative when it starts to fall.

Answer:

CODIS should be the answer

Answer and Explanation in one:

In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors. So basically, you could possibly demonstrate with a variety of examples. One example may be just sitting on a table, waiting to get rolled off. I know it sounds silly, so here's another example. You are holding a bow and arrow, you pull back the arrow, aiming at the target. You are holding the arrow, filling it with potential energy. When it is let go, it is kinetic energy, the act of an object in motion, slowly riding of it's potential energy.

Answer:

D

Explanation:

Gravity is a force that pulls objects down toward the ground. When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion.

A=9.8m/s^2 or the gravitational accelerational

If a stone dropped from the roof of a single-story building to the surface of the earth speeds up because of an almost constant force of gravity acting upon it,there4fore the correct answer is option D.

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another,

For any free-falling object, the acceleration due to gravity is constatant.

The correct response is option D if a stone thrown from the roof of a one-story structure to the surface of the ground accelerates due to an almost constant force of gravity acting upon it.

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The force acting on a moving charge is known as the magnetic force. The force acting on the charge will be 3.75 N.

Magnetic fields only exert a force on a moving electric charge. A moving charge generates a magnetic field. With an increase in charge and magnetic field strength, this force rises.

when charges have higher velocities, the force is stronger. However, the magnetic force is always perpendicular to the velocity.

Mathematically the force exerted on the charge will be

F=qvBsinα

F= force acting on the charge

v = velocity of charge

q = charge

F=qvBsinα

F=2.5×10⁻⁶×5.0×10³×3.0×10²

F=37.5 N

Hence The force acting on the charge will be 3.75 N.

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F = q V B sinα

Where F is the force applied to a moving charge.

V = charge velocity

q stands for charge.

α = angle between V and B directions

As a result, the moving charge is subjected to a force of 3.75 Newton.