a force applied to an object of mass m1 produces an acceleration of 3.60 m/s2 . the same force applied to a second object of mass m2 produces an acceleration of 1.60 m/s2 . what is the value of the ratio m1/m2?

The point between Earth and the moon where a space probe will experience no net force would be 384,400 km from Earth.

The point between Earth and the moon where a 50,000 kg space probe experience no net force is called the Lagrangian point. The fifth Lagrangian point (L5) is located about 60 degrees behind the moon, about 384,400 km from Earth. Therefore, the distance between the probe and the Earth is 384,400 km, which is the average distance between the Moon and Earth.

The Lagrangian point is a point in space where the gravitational forces of two major celestial bodies (such as Earth and the moon) or more celestial bodies balance the gravitational forces, allowing a third smaller body to remain in constant position relative to the larger bodies.

L5, the fifth Lagrangian point, is a Lagrangian point in the Earth-Moon system, located about 60 degrees behind the Moon. It is approximately 384,400 km away from Earth, the same as the average distance between Earth and the Moon. It is one of the stable equilibrium points of the Earth-Moon system, as the gravitational forces of the Earth and the Moon balance the centrifugal force acting on a spacecraft at this point.

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The frictional force exerted on the block when it slides along a rough surface is a non-zero force.

When the block comes to a stop, it can be concluded that the frictional force is equal in magnitude to the block's applied force but opposite in direction. This means that the frictional force is doing negative work since it is resisting the motion of the block. In other words, the frictional force is in the opposite direction of the motion and reduces the kinetic energy of the block until it stops.

The magnitude of the frictional force can be determined by the equation:

Ff = μFn, where Ff is the frictional force, μ is the coefficient of friction and Fn is the normal force.

The coefficient of friction is determined by the type of surfaces the block and the ground have. For example, if both the block and the ground are made of steel, the coefficient of friction would be higher than if the block was made of rubber and the ground was made of marble.

Therefore, when a block slides along a rough surface and comes to a stop, we can conclude that a non-zero frictional force is exerted on the block, which is equal in magnitude to the applied force but opposite in direction.

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The vertical component of the tension is 1,177.05 N while the horizontal component of the tension is 127.47 × 3.90² = 1,949.04 N.

The magnitude of the centripetal acceleration is 2.14 m/s².

A) The vertical component of the tension equals the weight it supports, which is the weight of the person plus the weight of the basket:

Weight = (82.0 kg + 45.0 kg) × 9.81 m/s²

Weight = 1,177.05 N

Therefore, the vertical component of the tension is 1,177.05 N.

To find the horizontal component of the tension, we can use the fact that the net force in the horizontal direction is zero when the basket is moving at a constant speed.

The only horizontal force is the component of the tension perpendicular to the radius, so:

The horizontal component of tension = centripetal force

Horizontal component of tension = (mass × centripetal acceleration)

Horizontal component of tension = (82.0 kg + 45.0 kg) × (v²/7.10 m)

Horizontal component of tension = 127.47 v² N

Setting these two components equal to each other gives:

1,177.05 N = 127.47 v² N

Solving for v gives:

v = 3.90 m/s

Therefore, the horizontal component of the tension is 127.47 × 3.90² = 1,949.04 N.

B) The centripetal acceleration is given by:

a = v²/r

a = (3.90 m/s)²/7.10 m

a = 2.14 m/s²

Therefore, the magnitude of the centripetal acceleration is 2.14 m/s².

C) The speed of the basket and its passenger is 3.90 m/s.

D) The time it takes the basket to make one complete circle is given by:

T = 2πr/v

T = 2π(7.10 m)/3.90 m/s

T = 12.9 s

Therefore, it takes the basket 12.9 s to make one complete circle.

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A torque of 3471.9 N·m is needed to produce an angular acceleration of 100 rpm/s in a flywheel with a diameter of 1.72 m and a mass of 902 kg.

First, let's convert the angular acceleration from revolutions per minute per second (rpm/s) to radians per second per second (rad/s²):

100 rpm/s = 100 × 2π/60 rad/s² ≈ 10.47 rad/s²

The moment of inertia of a flywheel can be calculated using the formula:

I = (1/2)mr²

where

m is the mass of the flywheel and

r is the radius (half of the diameter).

Thus, we have:

r = 1.72/2 = 0.86 m

m = 902 kg

I = (1/2) × 902 kg × (0.86 m)² ≈ 331.9 kg·m²

The torque (T) required to produce the desired angular acceleration (α) can be found using the formula:

T = I × α

T = 331.9 kg·m² × 10.47 rad/s² ≈ 3471.9 N·m

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When a ball is thrown at an upward angle, the vertical and horizontal components of velocity change in different ways. The vertical component of velocity decreases to a certain point before increasing again due to gravity. However, the horizontal component of velocity remains constant throughout the motion of the ball.

When a ball is tossed at an upward angle, the velocity has two components; vertical and horizontal components. The horizontal component is unaffected since there is no force acting on it.

The vertical component is influenced by the gravitational force acting on the ball. As the ball goes up, the vertical component of velocity decreases to zero. The maximum point is reached when the ball's velocity is zero. At this point, the ball stops going up and starts going down. As the ball falls, the vertical component of velocity increases in the opposite direction to the gravitational force acting on it.

Therefore, the vertical component of velocity changes as the ball is tossed at an upward angle. It increases, then decreases to zero at the top of its trajectory, and then increases again as the ball falls back to the ground. The horizontal component of velocity is constant throughout the motion of the ball because there is no force acting on it.

Hence, when a ball is tossed at an upward angle, the vertical and horizontal components of velocity change in different ways.

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The tension in the rope is 343.35 N.

To solve this question, we need to apply Newton's second law. In this scenario, the bucket is being lowered at a constant speed.

This means that the acceleration is zero. The forces acting on the bucket are gravity and tension.

Let's apply Newton's second law:ΣF = ma

Forces in the vertical direction:ΣF = 0

The forces acting on the bucket in the vertical direction are gravity (Fg) and tension (T).

Since the acceleration is zero, the net force must also be zero.

Therefore, the magnitude of the upward force (T) must be equal to the magnitude of the downward force (Fg).

Fg = mg

where m is the mass of the bucket and g is the acceleration due to gravity.

The force of tension can be calculated as follows:T = mg = (35.0 kg)(9.81 m/s²) = 343.35 N

The tension in the rope is 343.35 N.

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The pilot of an airplane notes that the compass indicates a heading due west. The airplane's speed relative to the air is 100 km/h. The air is moving in the wind at 31.0 km/h toward the north. The velocity of the airplane relative to the ground is: 104 km/h

The airplane's velocity relative to the ground is calculated by adding the velocity of the airplane relative to the air with the velocity of the air relative to the ground.

The velocity of the airplane relative to the ground is obtained by vector addition of the airplane's velocity relative to the air and the air's velocity relative to the ground. Given that the compass indicates a heading due west, the airplane's velocity relative to the air is 100 km/h towards the west.

The air is moving towards the north at 31.0 km/h, therefore the velocity of the air relative to the ground will be towards the north. The velocity of the air relative to the ground will be equal to 31.0 km/h towards the north.

To find the velocity of the airplane relative to the ground, we need to add the velocity of the airplane relative to the air to the velocity of the air relative to the ground.

Hence, we get the velocity of the airplane relative to ground = velocity of the airplane relative to air + velocity of air relative to ground. The velocity of the airplane relative to the ground = (100 km/h)2 + (31.0 km/h)2 = 104 km/h.

The velocity of the airplane relative to the ground is 104 km/h.

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Answer:

The is answer C

Explanation:

The electrons are always on the outside and the positive are in the inside the nucleus

and the neutron are in the inside.

Answer:

the correct option is C

Explanation:

in the orbitals that surrounds the nucleus .

thank you.

Depending on the spacing of the grating, there can be up to four bright spots seen on the screen behind the grating.


A grating is composed of multiple lines that are etched onto a surface, and when a light passes through these lines, it is split into its component wavelengths. Since the laser is 480 nm, the diffracted light will be composed of 480 nm light.

When light is shone through a grating, it diffracts and produces a pattern of bright spots and dark fringes on a screen placed behind the grating.

Depending on the spacing of the grating, there can be up to four bright spots seen on the screen behind the grating.

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Bohr developed an equation for calculating the energy levels of a hydrogen atom. Using this equation, the following can be determined:

The energy level of an electron

The angular momentum of an electron

The radius of the hydrogen atom's orbit

Around the nucleus of the hydrogen atom, the electrons move in circular orbits. Each of these orbits corresponds to a particular energy level.

Bohr's equation calculates these energy levels based on the electron's distance from the nucleus and its angular momentum.

Thus, by using Bohr's equation, we can determine the energy level of an electron, its angular momentum, and the radius of the hydrogen atom's orbit.

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Yes, adding too many fins on a surface can cause the overall heat transfer coefficient and heat transfer to increase.

This is because the presence of fins can increase the surface area available for heat exchange, allowing more heat to be transferred over a given period of time. Fins can also improve the convective heat transfer coefficient and turbulence levels of the surrounding fluid.
When adding fins to a surface, it is important to consider the fin spacing and height to ensure that the fins do not impede the flow of the surrounding fluid. For instance, if the fins are too close together, they can cause an increase in the pressure drop of the fluid and reduce the efficiency of the heat exchange. Likewise, if the fins are too high, they can block the flow of the fluid.
It is also important to consider the type of material used for the fins. Fin materials can affect the thermal conductivity of the fins, which in turn can influence the heat transfer rate. Furthermore, if the fins are made from a material that is not resistant to corrosion, the effectiveness of the fins may be reduced over time.
In summary, adding too many fins on a surface can cause the overall heat transfer coefficient and heat transfer to increase. It is important to consider the fin spacing, height, and material when determining the most efficient fin configuration for a given surface.

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The general process by which a large diffuse cloud of gas turns into a star and surrounding planets are known as: star formation.

The Star Formation process starts with a giant molecular cloud of gas and dust, where the gravitational forces act on the cloud and it collapses under its own gravity. This collapse results in a disc-like structure, which is also known as a protoplanetary disc, and has the potential to form planets.

The center of the disc gets hotter and denser, and eventually, nuclear fusion begins, resulting in the formation of a star. The protoplanetary disc contains a lot of dust and gas, and as the temperature increases, some of the minerals and elements present in the dust start to melt and then solidify, eventually forming small planetesimals, which aggregate to form the larger planets.

As the planets move around in the disc, they can migrate inward and outward, and some can collide and merge with others, thus forming even larger planets.

The remaining gas and dust in the disc are eventually swept up by the planets or blown away by the star's radiation, and the planets settle into stable orbits. This is the general process by which a large diffuse cloud of gas turns into a star and surrounding planets.

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The correct option is C. Both a particle and a wave accurately describe what light is. This is known as the wave-particle duality of light

Wave-particle duality is a fundamental concept in physics that describes the behavior of matter and energy at the atomic and subatomic scale. It states that matter and energy can exhibit both wave-like and particle-like behavior, depending on how they are observed or measured.

For example, light can be observed as both a wave and a particle, depending on the experiment. When it behaves as a wave, it exhibits characteristics such as diffraction, interference, and polarization. When it behaves as a particle, it exhibits characteristics such as energy and momentum. The wave-particle duality has significant implications for our understanding of the nature of reality and the fundamental laws of physics, and it has led to the development of many important technologies, such as lasers, transistors, and semiconductors.

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Complete Question: -

which choice accurately describes what light is? responses neither

A). a particle nor a wave neither

B). a particle nor a wave

C). both a particle and wave both a particle and a wave,

D). only a particle only a particle only a wave only a wave

The given statement, "surface currents flow vertically in the uppermost 400 meters of the water column," is false because surface currents flow horizontally in the uppermost 400 meters of the water column. They move water parallel to the surface, driven by factors such as wind and temperature differences.

Surface currents are driven by the wind, and they are characterized by movement across the surface of the water. The direction and intensity of surface currents are influenced by a variety of factors, including wind speed and direction, the shape of the coastline, and the rotation of the Earth. These currents are an essential component of the ocean circulation system and can have a significant impact on the climate and the distribution of marine life. They flow parallel to the water columns in the uppermost parts.

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If a 21.00-kg child slide from a height of 3.40 m above the bottom of the slide and her speed at the bottom is 2.30 m/s, the amount of energy lost due to friction is 644.18 J.

The potentiаl energy of аn object depends on the locаtion of the object from the bottom reference floor аnd the mаss of the object. The аmount of energy contаins by the object аt аny height is known аs the potentiаl energy of thаt object.

We are given:

The energy of the child at the upper end of the slide is,

[tex]E_{u}[/tex] = mgh

Substitute the values in the above equation

[tex]E_{u}[/tex] = 21 kg × 9.8 m/s2 × 3.40 m

= 699.72 J

The energy at the bottom of the slide is,

[tex]E_{b}[/tex] = [tex]\frac{1}{2}(mv^{2})[/tex]

Substitute the values in the above equation.

[tex]E_{b}[/tex] = [tex]\frac{1}{2}(21.2.30^{2})[/tex]

[tex]E_{b}[/tex] = 55.54 J

The energy lost due to friction is,

[tex]E_{f}[/tex] = [tex]E_{u}[/tex] - [tex]E_{b}[/tex]

Substitute the values in the above equation

[tex]E_{f}[/tex] = 699.72 - 55.54

[tex]E_{f}[/tex] = 644.18 J

Thus, the energy lost due to friction is 644.18 J.

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An emf source with a resistor and a capacitor are connected in series. as the capacitor charges, when the current in the resistor is 0.900 a. The magnitude of the charge on each plate of the capacitor will be: 0.900 A * t.

When an emf source with a resistor and a capacitor are connected in series, the current in the resistor will start decreasing as the capacitor charges up. When the current in the resistor is 0.900 A, the magnitude of the charge on each plate of the capacitor can be determined by the equation:

Q = I * t
where Q is the magnitude of the charge, I is current, and t is the time.

In this case, since the current is 0.900 A, the magnitude of the charge on each plate of the capacitor can be calculated by multiplying the current (0.900 A) by the time (t). The magnitude of the charge on each plate of the capacitor will therefore be 0.900 A * t.

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The energy consumption per hour of the heater is 9 kJ/hour and the energy consumption per hour of the electric motor is 3 kJ/hour.

Let's denote the energy consumption per hour of the heater as "h" and the energy consumption per hour of the electric motor as "m".

From the first piece of information, we can set up the equation:

2h + 4m = 25 (equation 1)

Similarly, from the second piece of information, we can set up another equation:

3h + 2m = 18 (equation 2)

We now have two equations with two unknowns, which we can solve using algebraic methods. Multiplying equation 2 by 2 and subtracting it from equation 1 multiplied by 3, we get:

(3h + 6m) - 2(3h + 2m) = 25(3) - 18(2)

Simplifying this expression, we get:

h = 9

Substituting this value of h into equation 2, we get:

3(9) + 2m = 18

Simplifying this expression, we get:

m = 3

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The moment of inertia of the bicycle wheel with radius of 0.304m and 50 spoke, rim with mass 2.50 kg for rim about the axle is 0.229 kg·m² , moment of inertia of any one spoke is 0.00186 kg·m² and  moment of inertia of the wheel, including the rim and all 50 spokes is 0.592 kg·m².

(a) The moment of inertia of the rim about the axle, we use the formula for the moment of inertia of a thin hoop.

We substitute the mass of the rim and the radius of the wheel into the formula and get the moment of inertia of the rim

The moment of inertia of the rim about the axle:

[tex]I_{rim} = MR^2[/tex]

where M is the mass of the rim and

R is the radius of the wheel.

Substituting the given values, we get:

[tex]I_{rim} = (2.50 kg) *(0.304 m)^2 = 0.229 kg*m^2[/tex]

Therefore, the moment of inertia of the rim about the axle is 0.229 kg·m².

(b) The moment of inertia of any one spoke, we use the formula for the moment of inertia of a long, thin rod rotating about one end.

We substitute the mass of the spoke and its length into the formula and get the moment of inertia of one spoke.

[tex]I_{spoke} = (1/3)ML^2[/tex]

where M is the mass of the spoke and

L is its length.

Substituting the given values, we get:

[tex]I_{spoke} = (1/3) *(0.0100 kg)*(2 * 0.304 m)^2= 0.00186 kg*m^2[/tex]

Therefore, the moment of inertia of any one spoke is 0.00186 kg·m².

(c) The total moment of inertia of the wheel, we use the parallel axis theorem.

The moment of inertia of the wheel about the center of mass is given by:

[tex]I_{center} = I_{rim} + 50*I_{spoke}[/tex]

Substituting the values we found in parts (a) and (b), we get:

[tex]I_{center} = 0.229 kg*m^2 + 50 * 0.00186 kg*m^2 = 0.324 kg*m^2[/tex]

The distance between the center of mass and the axle is equal to the radius of the wheel, so we can use the parallel axis theorem to find the total moment of inertia:

[tex]I_{total} = I_{center} + Md^2[/tex]

where M is the total mass of the wheel (rim plus spokes) and

d is the distance between the center of mass and the axle.

Substituting the given values, we get:

M = 2.50 kg + 50 × 0.0100 kg = 3.00 kg

d = 0.304 m

[tex]I_{total} = 0.324 kg*m^2 + (3.00 kg) *(0.304 m)^2= 0.592 kg*m^2[/tex]

Therefore, the total moment of inertia of the wheel, including the rim and all 50 spokes, is 0.592 kg·m².

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To determine the wavelength of a sound wave 1, the formula λ = v/f can be used, where λ represents the wavelength of the sound wave, v is the velocity of sound, and f is the frequency of the sound wave.

When sound waves propagate through a medium, they form a pattern of compressions and rarefactions that can be measured as sound waves.To test the theory with several different sounds, take note of the velocity and frequency of each sound. Here are the steps for determining wavelength of sound wave:1.

Measure the velocity of sound in a medium - this is constant in a given medium at a given temperature, so the value will be known.2. Determine the frequency of the sound wave. This is typically done with a microphone or other frequency-measuring device.3. Plug the values into the equation λ = v/f4. Solve for λ to find the wavelength of the sound wave.For example, suppose that the velocity of sound in a given medium is 343 meters per second, and the frequency of the sound wave is 440 hertz.

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In the following question, among the conditions given, The peak vertical ground reaction force (not resultant force) from the beginning of the measurement through leaving the ground in your spreadsheet is the highest vertical force.

Hence The peak vertical ground reaction force (not resultant force) from the beginning of the measurement through leaving the ground in your spreadsheet is the highest vertical force that the ground exerts on your body during the time period in question. so then, in order To calculate this, you need to examine your spreadsheet and look for the highest vertical force value present in the data.

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Empty beer can: mass 50g, length 12cm, radius 3.3cm. Moment of inertia found by subtracting mass of lid/bottom from mass of empty can, and using I=(1/2)mr² for a solid cylinder. Result: 1.7 x 10^-5 kg m².

An empty beer can has a mass of 50 g, a length of 12 cm, and a radius of 3.3 cm. Assume that the shell of the can is a perfect cylinder of uniform density and thickness. To find the moment of inertia of the can about the cylinder's axis of symmetry-

(a) Let the mass of the lid/bottom be m. The mass of the empty can is 50g.

Since the lid and bottom are identical in shape and mass, we can write that the total mass of the can is 2m + 50g.

Thus, the mass of the lid/bottom is m = (50g)/2 = 25g.

Therefore, the mass of the lid/bottom is 25g.

(b) The mass of the shell is the mass of the empty can minus the mass of the lid/bottom.

Therefore, the mass of the shell is

[tex]m_{shell} = m_{empty} - m_{lid/bottom} = 50g - 25g = 25g.[/tex]

(c) Moment of inertia of a solid cylinder of radius r and mass m about the axis of symmetry is given by

I = (1/2)mr²

The radius of the can is r = 3.3 cm = 0.033 m.

The length of the can is not needed to find the moment of inertia of the can about its axis of symmetry since the moment of inertia is independent of the length of the cylinder (as long as its mass and radius remain the same).

The mass of the shell is m_shell = 25g = 0.025 kg.

Using the formula for moment of inertia, we get

[tex]I = (1/2)mr² = (1/2)(0.025 kg)(0.033 m)² = 1.7 x 10^-5 kg m²[/tex]

Therefore, the moment of inertia of the can about its axis of symmetry is 1.7 x 10^-5 kg m².

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In the longitudinal wave ,B represents the phenomenon of rarefaction. Rarefaction refers to the region of a sound wave where the pressure of the medium is lower than its normal value.

Rarefaction is a term used to describe a decrease in the density or pressure of a substance, such as a gas or liquid. In the context of sound waves, rarefaction refers to the region of a sound wave where the pressure of the medium is lower than its normal value, causing the particles of the medium to be spread further apart than usual.

Sound waves are composed of regions of compression and rarefaction that alternate in a regular pattern as the wave travels through a medium. In a compressional (longitudinal) sound wave, the particles of the medium are pushed together in regions of compression, while they are spread apart in regions of rarefaction. These changes in pressure and density cause the wave to propagate through the medium.

In general, rarefaction can occur in any medium, not just in sound waves. For example, in a gas, rarefaction can be caused by a decrease in pressure, temperature or density. In a liquid, rarefaction can be caused by a decrease in pressure or density. Rarefaction waves can be observed in many natural phenomena, such as atmospheric pressure waves, seismic waves, and waves on the surface of water.

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The ground state energy of an infinite potential well with the width a decreases if we make the width smaller. The other energy levels also decrease but their energies are higher than the ground state energy.

This is because the energy levels of an infinite potential well are inversely proportional to the width of the well. That is, the energy levels increase as the width decreases and vice versa.

For an infinite potential well, the ground state energy is given by the expression:

$E_1=\frac{h^2}{8ma^2}$

Where, h is Planck’s constant

m is the mass of the particle

a is the width of the well.

This implies that as a decreases, the energy level of the ground state decreases as well. This can be seen in the graph below, which shows the variation of energy levels with the width of the well. The blue line corresponds to the ground state energy, which decreases as the width decreases.

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An object becomes electrically charged when there is a transfer of electrons between two objects. Electrons are negatively charged particles that orbit the nucleus of an atom. When two objects come into contact with each other, some electrons may move from one object to the other. The object that loses electrons becomes positively charged, while the object that gains electrons becomes negatively charged.

This transfer of electrons can also occur without direct contact between the objects. For example, if a charged object is brought close to a neutral object, the electrons in the neutral object may be attracted or repelled by the charged object. This can cause the electrons in the neutral object to move around, resulting in a separation of charges and the object becoming charged.

Another way an object can become charged is through the process of induction. If a charged object is brought near a neutral object, it can induce a separation of charges in the neutral object. This happens because the charged object creates an electric field that attracts or repels electrons in the neutral object. The result is a separation of charges, with one part of the object becoming positively charged and the other part becoming negatively charged.

There are two identical iron bars, one of which is a permanent magnet and the other unmagnetized. We can identify that: when the magnetized bar is brought near the other bar, it will stick to it, indicating that it is magnetized. The bar that does not stick is unmagnetized.

Iron bars are used to make permanent magnets by a process called magnetization. Permanent magnets are composed of atoms and aligned electrons that have magnetic properties. The other bar that is not magnetized does not have aligned electrons, so it will not attract other magnets as a magnetized bar would.

The direction of a magnetic field will change when a magnet is brought near it. The North Pole will attract the South Pole, and they will come together. The North Pole will repel the North Pole, and the South Pole will repel the South Pole. The magnetized bar will be attracted to the unmagnetized bar, and the unmagnetized bar will not be attracted to the magnetized bar.

As a result, when the magnetized bar is brought near the other bar, it will stick to it, indicating that it is magnetized. The bar that does not stick is unmagnetized. Thus, with the aid of two bars, one magnetized and the other unmagnetized, we can determine which is which.

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Answer:

L = L0 (1 - v^2 / c^2)^1/2

L0 is the proper length and L the distance measured by the space traveler

L = L0 (1 - .92^2)^1/2

L = L0 * .39 = 8.6 L-y * .39 = 3.4 L-y     as measured by space traveler

The potential difference between the two points in an electric field is 1 V.

Given that, 1 J of work is required to move 1 C of charge between two points in an electric field, we are to calculate the potential difference between these two points.

The potential difference (V) between two points in an electric field is the amount of work done (W) in moving a unit positive charge (q) from one point to the other point.

Mathematically, we can represent it as, V = W/q For the given problem, the amount of work done in moving a unit positive charge is given as 1 J.

So we can write it as, W = 1 J Also, the amount of charge moved is 1 C. So we can write it as, q = 1C

Now substituting these values in the above expression for potential difference (V), we get, V = W/q = 1 J/1 C = 1 V.

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A portable CD player uses 7.5mA of current at a potential difference of 3.5V.  Since it is running for 35 seconds, the total energy consumed in that time is  calculated by the product of potential difference, current and time consumed and it is solved as 918.75mJ.

The amount of energy used by the portable CD player can be calculated using the formula:

E = VIt

where E is the energy, V is the potential difference, I is the current and t is the time.

The portable CD player uses a current of 7.5 mA at a potential difference of 3.5 V.

Thus, the energy used by the player in 35 seconds can be calculated as follows:

[tex]E = VIt\\ = 3.5 V \times 7.5 mA \times35 s \\= 918.75 mJ[/tex]

Therefore, the portable CD player uses 918.75 mJ of energy in 35 seconds.

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The acceleration of the two blocks is g/4.

Tension force in the left section of the string is 5/4 Mg

Tension force in the right section of the string is 3/4 Mg

Angular acceleration of the pulley is 0.

a. The acceleration of the two blocks can be found by applying Newton's second law to each block. For Block A, the force equation is:

T - Mg = Ma

where T is the tension force in the string, M is the mass of Block A, g is the acceleration due to gravity, and a is the acceleration of Block A. For Block B, the force equation is:

3Mg - T = 3Ma

where T is the tension force in the string and a is the acceleration of Block B. Since the string is assumed to be light and inextensible, the tension force in both sections of the string is the same.

The two equations can be solved simultaneously to obtain the acceleration: a = g/4

b. To find the tension force in the left section of the string, we can use the force equation for Block A:

T - Mg = Ma

Substituting the value of acceleration we obtained in part a:

T = 5/4 Mg

c. To find the tension force in the right section of the string, we can use the force equation for Block B:

3Mg - T = 3Ma

Substituting the value of acceleration we obtained in part a, and the value of T we obtained in part bt:

T = 3/4 Mg

d. To find the angular acceleration of the pulley, we can use the torque equation:

Iα = Στ

where I is the moment of inertia of the pulley, α is the angular acceleration, and Στ is the net torque acting on the pulley.

The tension force in the string exerts a torque on the pulley, given by:

τ = TR

where R is the radius of the pulley. Since the tension force is the same on both sides of the pulley, the net torque is zero. Thus, we have:

Iα = 0 which implies that the angular acceleration of the pulley is zero.

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The centrifuge must rotate at approximately 54959 rpm to produce an acceleration of 115000 g's at a distance of 8.50 cm from the axis of rotation.

To solve this problem, we can use the formula for centrifugal acceleration:

a = (r * w^2) / g

where a is the desired acceleration in units of g's, r is the distance of the particle from the axis of rotation, w is the angular velocity of the centrifuge in radians per second, and g is the acceleration due to gravity (approximately 9.81 m/s^2).

First, we need to convert the distance from centimeters to meters:

r = 8.50 cm = 0.085 m

Next, we can rearrange the formula to solve for the angular velocity w:

w = sqrt((a * g) / r)

Substituting the given values, we get:

w = sqrt((115000 * 9.81) / 0.085)

w = 5758.6 radians per second

Finally, we can convert the angular velocity from radians per second to revolutions per minute (rpm):

1 revolution = 2π radians

1 minute = 60 seconds

w (in rpm) = (w / 2π) * 60

w (in rpm) = (5758.6 / (2π)) * 60

w (in rpm) ≈ 54959

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