A copper wire of length 2.0 m, cross sectional area 7.1 ´ 10-6 m2 and Young's modulus 11 ´ 1010 N/m2 has a 200-kg load hung on it. What is its increase in length?

Answers

Answer 1

The increase in length of the copper wire is 5.4 x 10^-4 meters.

Calculate the increase in length.

We can use the formula for the stress-strain relationship:

stress = force/area

strain = change in length / original length

Young's modulus is defined as:

Young's modulus = stress/strain

Therefore, we can rearrange these equations to solve for the change in length:

change in length = (force * length) / (area * Young's modulus)

Plugging in the given values, we get:

change in length = (200 kg * 9.81 m/s^2 * 2.0 m) / (7.1 x 10^-6 m^2 * 11 x 10^10 N/m^2)

= 5.4 x 10^-4 m

Therefore, the increase in length of the copper wire is 5.4 x 10^-4 meters (or 0.54 millimeters).

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Related Questions

What is the intensity of sound from a band with a sound level of 120 dB? (I0 = 10-12 W/m2)

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The intensity of sound from a band with a sound level of 120 dB is 1 W/m2.

The intensity of sound is defined as the amount of energy that is transmitted through a unit area per unit time. It is usually measured in watts per square meter (W/m2). The sound level of a band is usually measured in decibels (dB), which is a logarithmic unit of measurement that expresses the ratio of the sound pressure level to a reference value.
To determine the intensity of sound from a band with a sound level of 120 dB, we need to use the following formula:
[tex]I = I0 \times  10^{(L/10)[/tex]
Where I0 is the reference intensity (10-12 W/m2), L is the sound level in decibels (120 dB), and I is the intensity of sound in watts per square meter (W/m2).
Plugging in the values, we get:
[tex]I = 10^-12 \times  10^{(120/10)\\I = 10^-12 \times  10^{12[/tex]
I = 1 watt per square meter (W/m2)
Therefore, the intensity of sound from a band with a sound level of 120 dB is 1 W/m2. This means that the band is producing a very loud sound that can cause hearing damage if exposure is prolonged. It is important to protect your ears when attending loud concerts or events by wearing earplugs or earmuffs.

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If you catch an egg of mass m that is heading toward your hand at speed v what magnitude momentum change does it undergo? Hint: The egg is at rest after you catch it.

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When you catch an egg of mass m that is heading toward your hand at speed v, the egg undergoes a momentum change equal to its initial momentum. The momentum of the egg changes by an amount equal to its initial momentum, which is mv.

Momentum is defined as the product of mass and velocity and is a vector quantity with magnitude and direction.

Since the egg is initially moving with a velocity of v, its momentum is given by

p = mv.

When you catch the egg, its velocity drops to zero, and hence its momentum also drops to zero.

Therefore, the magnitude of the momentum change is equal to the initial momentum of the egg, which is given by

|Δp| = |p_f - p_i| = |-mv| = mv

In other words, when you catch the egg, the momentum of the egg changes by an amount equal to its initial momentum, which is mv.

This momentum change is also equal to the force that you exert on the egg to bring it to a stop, as per Newton's second law of motion (F = Δp/Δt).

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The electric potential increases from 52 V to 331 V from the bottom plate to the top plate of a parallel-plate capacitor. A −4 ×10-3 C charge is moved from the bottom plate to the top plate. What is the magnitude of the change in potential energy of the charge?

Answers

The magnitude of the change in potential energy of the charge is 1.116 J.

This can be calculated using the equation:

ΔPE = qΔV

where ΔPE is the change in potential energy, q is the charge moved, and ΔV is the change in electric potential.

In this case, q = -4 ×10-3 C (negative because the charge is moved from the bottom plate to the top plate) and ΔV = 331 V - 52 V = 279 V (the potential difference between the plates).

Substituting these values into the equation, we get:

ΔPE = (-4 ×10-3 C) x (279 V)
ΔPE = -1.116 J

Therefore, the magnitude of the change in the potential energy of the charge is 1.116 J.

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Between 0° and 4°C, the volume coefficient of expansion for water:
a. is positive.
b. is zero.
c. is becoming less dense.
d. is negative.

Answers

It’s A because I took the test

T/F The sign of the angular velocity vector is always in the direction of the rotation

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True, the sign of the angular velocity vector is always in the direction of rotation. Angular velocity is a vector quantity that describes the rate of change of angular position of an object with respect to time. The direction of the angular velocity vector is determined by the right-hand rule, where the direction of rotation is the direction of the thumb and the direction of the angular velocity vector is the direction of the curled fingers around the thumb.

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As you watch the tumbler in a clothes dryer spin clockwise at a constant speed, what is the direction of the angular velocity?

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The direction of angular velocity is given by the Right hand-thumb rule. The tumbler in the clothes dryer spin in a clockwise direction, then the direction of angular velocity is downwards.

The direction of angular velocity is given by the Right hand-thumb rule. This rule states that if the curly fingers point in the clockwise direction, then the angular velocity points downward.  If the curly fingers point in the direction of an anti-clockwise direction, then the thumb fingers point upwards.

Thus, the if the tumbler in a clothes dryer spin in a clockwise direction, angular velocity is in a downward direction.

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factors that affect friction between two surfaces include the?question 5 options:amount of force pressing the two forces togethersmoothness of the two surfacesweight of the object resting on topall of the above

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All of the above factors affect friction between two surfaces, including the amount of force pressing the surfaces together, the smoothness of the surfaces, and the weight of the object resting on top.

Several things affect friction between two surfaces. The amount of pressure exerted on the two surfaces is important because it influences the normal force, which in turn impacts the frictional force. Friction is also influenced by the smoothness or roughness of the surfaces because rougher surfaces provide more interlocking points, which raise friction.

Friction is impacted by weight because increased normal force leads to increased frictional force when an object is lying on top of two surfaces. In conclusion, each of these variables influences how much friction there is between two surfaces.

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you may have observed that the amplitude of the spring oscillations in part 2 decreases over time. what causes the oscillations to become damped (decreased) over time?

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The decrease in the amplitude of spring oscillations over time is caused by damping. Damping refers to the process of removing energy from the system, which results in a decrease in the amplitude of oscillations. There are two types of damping: viscous damping and non-viscous damping.

Viscous damping occurs due to the presence of a fluid, such as air or water, which resists the motion of the spring. The fluid absorbs the energy of the spring and converts it into heat, which results in a decrease in the amplitude of oscillations. Non-viscous damping occurs due to the presence of external forces, such as friction or resistance, which also remove energy from the system.

In the case of spring oscillations, damping can be caused by a variety of factors, such as air resistance, friction between the spring and its surroundings, or internal friction within the spring itself. As the oscillations continue, the energy of the system is gradually dissipated, leading to a decrease in the amplitude of oscillations.

In summary, the damping of spring oscillations occurs due to the removal of energy from the system, which can be caused by a variety of factors. As the energy of the system is dissipated, the amplitude of oscillations gradually decreases over time.

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In order to overcome a surface tension of a fluid, a force of 1.32 x 10-2 N is required to lift a wire ring of circumference 12.0 cm. What is the surface tension of the fluid?

Answers

The surface tension of the fluid is 0.11 N/m.

Surface tension is the tendency of liquid surfaces at rest to shrink into the minimum surface area possible. Surface tension is what allows objects with a higher density than water such as razor blades and insects to float on a water surface without becoming even partly submerged.

To calculate the surface tension of the fluid, we can use the formula:
Surface tension = Force / Circumference

Plugging in the given values, we get:
Surface tension = 1.32 x 10⁻² N / 12.0 cm

We need to convert the circumference to meters, so:
Surface tension = 1.32 x 10⁻² N / (0.12 m)

Surface tension = 0.11 N/m

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what is translational equilibrium? What does this mean in terms of movement of an object?

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The object will continue to move in a straight line at a constant velocity unless acted upon by an external force.

Translational equilibrium refers to a state where an object is not experiencing any net force that is causing it to move in a particular direction. This means that the object is either at rest or moving with a constant velocity. In terms of movement, it means that the object's motion is stable and not changing in direction or speed. In terms of movement, this means that the object will either remain at rest or continue moving at a constant velocity, as there is no unbalanced force to cause any change in its motion.

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If the sound level of a heavy truck is 85 dB at a distance of 16 m, what is the sound level at a distance of 4 m from the truck?

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The sound level at 4 m from the truck would be 97 dB.

The sound level of a heavy truck at a distance of 85 dB at a distance of 16 m is considered to be quite loud, and it can have adverse effects on the human ear. When we move closer to the source of the sound, the intensity of the sound increases, which in turn increases the sound level.

To calculate the sound level at a distance of 4 m from the truck, we can use the inverse square law formula, which states that the intensity of sound is inversely proportional to the square of the distance from the source.

Thus, if the sound level of a heavy truck is 85 dB at 16 m, then at 4 m, the distance is 1/4th of 16 m. Hence, the sound level will increase by 4^2 = 16 times.

Therefore, the sound level at 4 m from the truck would be 85 + 10*log10(16) = 97 dB.

In conclusion, as we move closer to the source of the sound, the sound level increases, and it can have negative effects on the human ear. It is essential to take necessary precautions like wearing ear protection to avoid any hearing damage when exposed to loud sounds for prolonged periods.

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A large stone is resting on the bottom of the swimming pool. The normal force of the bottom of the pool on the stone is equal to the:

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The normal force of the bottom of the pool on the stone is equal to the weight of the stone.

According to Newton's third law of motion, every action has an equal and opposite reaction. In this case, the stone is exerting a force on the bottom of the pool due to its weight. As a result, the bottom of the pool exerts an equal and opposite force on the stone, which is the normal force. Since the stone is at rest, the normal force is equal in magnitude and opposite in direction to the force of gravity acting on the stone, which is the weight of the stone.

Therefore, the normal force of the bottom of the pool on the stone is equal to the weight of the stone.

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a golf ball has been hit off of the tee at an angle of elevation of 30 degrees and an initial velocity of 160 ft/sec. using the helpful parametric models for projectile motion answer the following questions. a.) how long is the ball in the air (hang time)? b.) what is the maximum height of the ball? c.) how far, horizontally, does the ball travel in the air? (round to the nearest foot)

Answers

The ball is in the air for approximately 4.15 seconds.

What is Projectile Motion?

Projectile motion refers to the motion of an object that is launched or thrown into the air and then moves under the influence of gravity alone. The object follows a curved path, known as a trajectory, as it moves through the air.

To solve this problem, we can use the following parametric equations for the x and y positions of the golf ball as a function of time:

x(t) = V0 * cos(theta) * t

y(t) = V0 * sin(theta) * t - (1/2) * g *[tex]t^{2}[/tex]

where V0 is the initial velocity of the golf ball, theta is the angle of elevation, and g is the acceleration due to gravity.

a.) The ball will be in the air until it hits the ground, which occurs when y = 0. Solving the y equation for t when y = 0 gives:

0 = V0 * sin(theta) * t - (1/2) * g * [tex]t^{2}[/tex]

t = (2 * V0 * sin(theta)) / g

Plugging in the given values, we get:

t = (2 * 160 ft/sec * sin(30 deg)) / (32.2 ft/[tex]sec^{2}[/tex]) ≈ 4.15 sec

Therefore, the ball is in the air for approximately 4.15 seconds.

b.) The maximum height of the ball occurs when the y velocity is zero. Solving the y equation for t when vy = 0 gives:

0 = V0 * sin(theta) - g * t

t = V0 * sin(theta) / g

Plugging in the given values, we get:

t = 160 ft/sec * sin(30 deg) / 32.2 ft/[tex]sec^{2}[/tex] ≈ 2.03 sec

Substituting this time back into the y equation gives the maximum height:

y_max = V0 * sin(theta) * t - (1/2) * g * [tex]t^{2}[/tex]

y_max = 160 ft/sec * sin(30 deg) * 2.03 sec - (1/2) * 32.2 ft/[tex]sec^{2}[/tex] * (2.03 sec)^2

y_max ≈ 116.5 ft

Therefore, the maximum height of the ball is approximately 116.5 feet.

c.) The horizontal distance traveled by the ball is given by the x position at the time the ball hits the ground, which is:

x = V0 * cos(theta) * t

x = 160 ft/sec * cos(30 deg) * 4.15 sec

x ≈ 561 ft

Therefore, the ball travels approximately 561 feet horizontally in the air.

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What is the life cycle of a metal resource?

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A metal's useful life is the full period from when it is first mined until it evaporates, or is finely scattered, in the environment and can no longer be used economically. The most durable metals are iron and steel alloys, which may be used for 150 years on average.

The great efficiency of the industrial processes used to treat these metals and high recycling rates, according to the researchers, are the main causes of this. Even if the lifespan of precious metals like gold and silver and non-ferrous metals like copper and aluminium is substantially shorter, it is still greater than 50 years. Contrarily, metals that are technology-specific and occasionally essential—i.e., scarce—only last for around twelve years in the economic cycle. Examples of this diverse category of basic minerals include cobalt and indium. The most durable metals are iron.

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one of the harmonic frequencies of tube a with two open ends is 325 hz. the next- highest harmonic frequency is 390 hz. (a) what harmonic frequency is next highest after the harmonic frequency 195 hz? (b) what is the number of this next-highest harmonic? one of the harmonic frequencies of tube b with only one open end is 1080 hz. the next-highest harmonic frequency is 1320 hz. (c) what harmonic frequency is next highest after the harmonic frequency 600 hz? (d) what is the number of this next- highest harmonic?

Answers

(a) The difference between the two given harmonic frequencies for tube A is 390 Hz - 325 Hz = 65 Hz. This difference is the fundamental frequency of the tube. The next highest harmonic frequency after 195 Hz will be 195 Hz + 65 Hz = 260 Hz.

(b) The fundamental frequency is 65 Hz, so the harmonic number for 260 Hz can be found by dividing the frequency by the fundamental frequency: 260 Hz / 65 Hz = 4. Thus, the next-highest harmonic is the 4th harmonic.

(c) The difference between the two given harmonic frequencies for tube B is 1320 Hz - 1080 Hz = 240 Hz. This difference is three times the fundamental frequency of tube B, as it has only one open end. The fundamental frequency is 240 Hz / 3 = 80 Hz. The next highest harmonic frequency after 600 Hz will be 600 Hz + 240 Hz = 840 Hz.

(d) Since tube B has only one open end, it will only produce odd harmonics. The fundamental frequency is 80 Hz, so the harmonic number for 840 Hz can be found by dividing the frequency by the fundamental frequency: 840 Hz / 80 Hz = 10.5. Since it is an odd harmonic, multiply by 2 and subtract 1: (10.5 * 2) - 1 = 20. Thus, the next-highest harmonic is the 20th harmonic.

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When you talk, you change the shape of the vocal tract to form different speech sounds. What happens to the frequency spectrum when the shape of the vocal tract is changed from a neutral configuration?

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The resulting changes to the frequency spectrum of the sound wave are what allow us to distinguish between different speech sounds.

When the shape of the vocal tract is changed from a neutral configuration, the frequency spectrum of the speech sound is altered. This is because the different resonant frequencies of the vocal tract change as the shape of the tract changes.

The vocal tract acts as a resonator, meaning it amplifies certain frequencies of the sound produced by the vocal cords while attenuating others. When the shape of the vocal tract changes, the resonant frequencies also change, resulting in a different pattern of amplification and attenuation of frequencies in the sound wave. This results in different frequency components being emphasized or suppressed, leading to different sounds.

For example, when the shape of the vocal tract is changed to produce a higher frequency sound like the "ee" sound in "sheep", the resonant frequency of the vocal tract is increased, causing higher frequency components of the sound wave to be amplified more strongly. Similarly, when the shape of the vocal tract is changed to produce a lower frequency sound like the "ah" sound in "father", the resonant frequency of the vocal tract is lowered, causing lower frequency components of the sound wave to be amplified more strongly.

Overall, changing the shape of the vocal tract is a critical part of producing different speech sounds, and the resulting changes to the frequency spectrum of the sound wave are what allow us to distinguish between different speech sounds.

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An 84.7 kg cross country skier on snow has uk = 0.131. With how much force must he push to move at a constant speed?

Answers

The skier must push with a force of 110.5 N to move at a constant speed on the snow.

To calculate the force required to move the cross country skier at a constant speed, we need to use the formula:

F = m * g * uk

where F is the force required, m is the mass of the skier, g is the acceleration due to gravity (9.81), and uk is the coefficient of kinetic friction between the skis and the snow.

Substituting the given values, we get:

F = 84.7 kg * 9.81 * 0.131

= 110.5 N

The coefficient of kinetic friction (uk) is a dimensionless quantity that represents the amount of friction between two surfaces in relative motion. In this case, it represents the friction between the bottom of the skis and the snow.

The value of uk depends on several factors, such as the roughness of the surfaces, the temperature, and the pressure.

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9. In a Jules Verne novel, a piece of ice is shaped into a magnifying lens to focus sunlight to start a fire. Is that possible?

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In a Jules Verne novel, a piece of ice is shaped into a magnifying lens to focus sunlight to start a fire, this is possible to use a piece of ice as a magnifying lens to start a fire, but it requires specific conditions and techniques.

Firstly, the ice must be clear and free of impurities, as these would obstruct and scatter the light, making it difficult to focus. Secondly, the ice must be properly shaped into a convex lens that can concentrate sunlight onto a small area, similar to a standard magnifying glass. This can be done by carving and polishing the ice with a knife or other tools until it forms a smooth, rounded surface.

Once the ice lens is prepared, it is crucial to position it correctly between the sun and the kindling material. The focal point, where the sunlight is most concentrated, must be aimed precisely at the material to generate enough heat to ignite it. Dry leaves, grass, or other fine materials are the most effective for starting fires, as they catch fire more easily than thicker, denser substances. In conclusion, it is indeed possible to use an ice lens to focus sunlight and start a fire, as portrayed in Jules Verne's novel. However, this technique requires clear ice, proper shaping, and accurate positioning to be successful.

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if a beam of 10 kev x rays illuminates a sample, what angles will give diffraction maxima of the first, second and third order? express your answers in degrees separated by commas.

Answers

The angles of diffraction maxima for a beam of 10 keV X-rays on a crystal can be calculated using Bragg's Law.

How to calculate angles for diffraction maxima using Bragg's Law?

Assuming the sample is a crystal and using Bragg's Law, the angles that give diffraction maxima can be calculated using the equation:

nλ = 2d sinθ

Where n is the order of diffraction (1, 2, 3, etc.), λ is the wavelength of the X-rays (in this case, 10 keV corresponds to a wavelength of approximately 0.124 nm), d is the distance between the crystal lattice planes, and θ is the angle of diffraction.

Using typical values for d-spacing in crystalline materials, we can calculate the angles for the first three orders of diffraction:

For first order (n=1): sinθ = λ/2d = 0.124/2d

θ = [tex]sin^-^1[/tex] (0.124/2d) = angle in degrees

For second order (n=2): sinθ = 2λ/2d = 0.124/ d

θ = [tex]sin^-^1[/tex] (0.124/d) = angle in degrees

For third order (n=3): sinθ = 3λ/2d = 0.372/d

θ = [tex]sin^-^1[/tex] (0.372/d) = angle in degrees

The exact values of the angles depend on the value of d for the crystal, which varies for different materials.

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Angular velocity derived from angular position: Disk rotating around its axis like a merry go round. What is the angular position of the disk at t= −2.00 s? Theta(t) equation is given.

Answers

The angular position of the disk at t = -2.00 s is 14.00 radians.

To determine the angular position of the disk at t = -2.00 s, we need to use the given equation for theta(t) and substitute t = -2.00 s.

Assuming that the angular position is given in radians, the equation for theta(t) can be written as:

θ(t) = [tex]3t^2[/tex]- 5t + 2

Substituting t = -2.00 s, we get:

θ(-2.00) = [tex]3(-2.00)^2[/tex] - 5(-2.00) + 2

= 3(4.00) + 10.00 + 2

= 14.00

Which means that at t = -2.00 s, the disk's angular position is 14.00 radians.

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fill in the blank At equilibrium at 1200°C, [I2] = 9.5 ×10-2 M
and [I] = 3.2 × 10-2 M.
What is the value of Keq for this system?
In this system, the equilibrium lies to the _____,
and the reaction favors the _______ 1.1 × 10-2
left
reactants

Answers

The equilibrium constants (Keq) of this system at equilibrium at 1200 °C is 1.1 10-2. In such a system, the reaction favours the reactants and the equilibrium is to the left.

Why is the equilibrium constant important?

By comparing the ratio between reactant to end product in a chemical reaction, it is possible to compute the equilibria constant, which is used to predict chemical behaviour. At equilibrium, the velocity of the reaction moving forward is equal to the velocity of the reaction moving backward.

The equilibrium constant is where?

Once a process has reached equilibrium, the concentrations or each component are measured in order to determine the numerical significance for the equilibrium constant. Calculated is the ratio of reactant concentrations to product concentrations.

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an electron is shot through a spot somewhere between the ends of a horseshoe magnet. the electron is repelled by both poles, and therefore is turned back. speed is increased. is unaffected by the field. direction is changed. is attracted to one of the poles, and repelled by the other.

Answers

Based on the given scenario, the electron would be affected by the magnetic field of the horseshoe magnet.

As it is shot through the spot between the ends of the magnet, it would experience a repulsive force from both poles.

This would cause its direction to change and it would be turned back.

The increase in speed would depend on the strength of the magnetic field and the distance between the electron and the poles.

Ultimately, the electron would not be unaffected by the field, but rather would experience a force that changes its trajectory.

It is not necessarily attracted to one pole and repelled by the other, but rather experiences a repulsive force from both poles.

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Does the total momentum change differ if you catch the egg more slowly or is it the same?

Answers

The total momentum change of an egg being caught would not differ based on the speed at which it is caught. This is due to the law of conservation of momentum, which states that in a closed system, the total momentum before an event must be equal to the total momentum after the event.

In this case, the system includes both the egg and the catcher. When the egg is falling, it has a certain amount of momentum due to its mass and velocity.

When it is caught, the catcher applies an equal and opposite force, resulting in a change in the egg's momentum. However, the total momentum of the system (egg and catcher) must remain constant.

Thus, the total momentum change would be the same regardless of the speed at which the egg is caught.

It is important to note that this assumes a perfectly closed system with no external forces acting on the egg or catcher. In reality, there may be slight variations due to factors such as air resistance and the force applied by the catcher.

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a car is towing a 140 kg refrigerator across a warehouse yard, pulling it by a rope tied around it. the rope makes a 25 degree angle relative to the horizontal. the refrigerator has a 0.25 coefficient of kinetic friction with the ground. the car drives at a constant speed across the yard. what is the magnitude of the work done by friction on the refrigerator when the car drags it across the 10 meter yard? please give the positive magnitude and do not include units in your answer.

Answers

The work done by the friction on the refrigerator when the car drags it across the 10-meter yard is 3.499 kJ.

Work done equals the product of force and displacement. It defines the energy transferred to the system by the means of force. Work done = force × displacement. When the body is in motion, it experiences a frictional force. The frictional force is the product of normal force and the coefficient of kinetic friction(μk).

From the given,

mass of the refrigerator (m) = 140kg

distance travelled d = 10m

co-efficient of kinetic friction (μk) = 0.25

form an angle θ = 25°

Force = (μk) × Fₙ   (Fₙ represents the normal force )

Fₙ = m×g = 140 × 10 = 1400 N ( g represents acceleration due to gravity)

Force = (μk) × Fₙ

         =  0.25 × 1400

         = 350 N

Force F = 350 N

The work done,

W = F×d

   = F×d×cosθ

   = 350×10×(cos(25°))

   =  3.499kJ

Thus, the work done by friction is W = 3.499kJ.

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. consider an electron accelerated in a vacuum between two parallel plates 25 cm apart. if you apply 5000 v across the plates, and the electron starts at rest on the surface of one plate, what is its velocity when it hits the opposite plate?

Answers

To solve this problem, we need to use the equation for the acceleration of an electron in an electric field:

a = eE/m

Where a is the acceleration, e is the charge of the electron, E is the electric field strength, and m is the mass of the electron.

In this case, we know that the distance between the plates is 25 cm, so the electric field strength can be calculated as:

E = V/d = 5000/0.25 = 20000 V/m

Now we can calculate the acceleration of the electron:

a = (1.6 x 10^-19 C) x (20000 V/m) / (9.11 x 10^-31 kg) = 3.53 x 10^14 m/s^2

Next, we need to use the equations of motion to find the velocity of the electron when it reaches the opposite plate. Since the electron starts at rest, we can use the following equation:

v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity (which is zero), a is the acceleration we just calculated, and s is the distance between the plates (25 cm = 0.25 m).

Substituting in the values, we get:

v^2 = 0 + 2 x (3.53 x 10^14 m/s^2) x 0.25 m = 4.41 x 10^14 m^2/s^2

Taking the square root of both sides gives us:

v = 2.10 x 10^7 m/s

So the velocity of the electron when it hits the opposite plate is approximately 21 million meters per second.
Hi! To find the velocity of the electron when it hits the opposite plate, we need to consider the electric potential difference (5000 V), the distance between the plates (25 cm), and the charge and mass of the electron.

First, convert the distance to meters: 25 cm = 0.25 m.

Next, use the formula for electric potential energy: PE = qV, where PE is potential energy, q is the charge of the electron, and V is the electric potential difference.

For an electron, q = -1.6 x 10^-19 C. Therefore, PE = (-1.6 x 10^-19 C)(5000 V) = -8 x 10^-16 J.

Since the electron starts at rest, all its potential energy will be converted to kinetic energy: KE = (1/2)mv^2, where m is the mass of the electron, and v is its velocity.

The mass of an electron is 9.11 x 10^-31 kg. Equating potential energy to kinetic energy:

-8 x 10^-16 J = (1/2)(9.11 x 10^-31 kg)v^2.

Solve for v: v = sqrt((-2 * 8 x 10^-16 J)/(9.11 x 10^-31 kg)) ≈ 1.32 x 10^7 m/s.

So, the electron's velocity when it hits the opposite plate is approximately 1.32 x 10^7 m/s.

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13.41 A 15 m long garden hose has an inner diameter of 2.5 cm. One end is connected to a spigot; 20 degree C water flows from the other end at a rate of 1.2 L/s. What is the gauge pressure at the spigot end of the hose?A 1900 paB 2700 paC 4200 paC 5800 paE 7300 pa

Answers

Answer: the gauge pressure at the spigot end of the hose is approximately 4204 Pa, which is closest to option C: 4200 Pa.

Explanation: We can use Bernoulli's equation to determine the gauge pressure at the spigot end of the hose. Bernoulli's equation relates the pressure, velocity, and height of a fluid in a horizontal pipe. For an incompressible fluid (like water) flowing through a horizontal pipe of constant diameter, Bernoulli's equation can be written as:

P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2

where:

P1 = pressure at the spigot end (unknown)

v1 = velocity at the spigot end (0 m/s, since the water is not moving initially)

P2 = pressure at the end of the hose (unknown)

v2 = velocity at the end of the hose (1.2 L/s = 0.0012 m^3/s ÷ (π/4)(0.025 m)^2 = 1.215 m/s)

ρ = density of water at 20°C = 998 kg/m^3

We can simplify the equation by assuming that the pressure at the end of the hose (P2) is atmospheric pressure (101325 Pa), since the water is flowing out into the open air. Plugging in the values and solving for P1, we get:

P1 = P2 + (1/2)ρv2^2 - (1/2)ρv1^2

= 101325 Pa + (1/2)(998 kg/m^3)(1.215 m/s)^2 - (1/2)(998 kg/m^3)(0 m/s)^2

≈ 4204 Pa

Therefore, the gauge pressure at the spigot end of the hose is approximately 4204 Pa, which is closest to option C: 4200 Pa.

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Suppose a Uranium 238 nucleus "swallows" a neutron and then decays not by fission but by beta decay, emitting an electron and an anti-neutrino. Which nucleus remains after this decay?Answera. 239Pub. 238Npc. 239Npd. 238Pae. 239U

Answers

When a Uranium 238 nucleus absorbs a neutron, it becomes Uranium 239. The correct answer is e. 239U.

The Uranium 239 nucleus then undergoes beta decay, where a neutron in the nucleus is converted into a proton, releasing an electron and an anti-neutrino. This results in the nucleus having one more proton, making it a new element. Therefore, the final nucleus after this decay is 239U.

When a Uranium 238 nucleus "swallows" a neutron and then decays via beta decay, emitting an electron and an anti-neutrino, the nucleus that remains after this decay is 239Pu.
Here's the step-by-step explanation:
1. The Uranium 238 nucleus captures a neutron, becoming Uranium 239 (U-239).
2. The U-239 undergoes beta decay, in which a neutron is converted into a proton, releasing an electron and an anti-neutrino.
3. This process results in the formation of a Plutonium 239 (Pu-239) nucleus, as there is now an additional proton.
So, the correct answer is a. 239Pu.

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Given that the frequency of a rotating bicycle tire is 15 Hz, find the angular frequency.

Answers

Answer: 30π rad

Explanation:

a student is doing experiments involving magnetic fields with the two long wires shown above, which are in the plane of the page. the student connects the ends of the left wire to a battery, creating a current in the wire toward the bottom of the page. what is the direction of the magnetic field at the location of the right wire?

Answers

The direction of the magnetic field at the location of the right wire is into the plane of the page.

To determine the direction of the magnetic field at the location of the right wire, we need to consider the following terms: magnetic field, current, and direction.

Step 1: Identify the direction of the current in the left wire.

The current flows toward the bottom of the page, as stated in the question.

Step 2: Apply the right-hand rule to find the direction of the magnetic field.

Place your right thumb in the direction of the current (downward) and curl your fingers. Your fingers will curl in the direction of the magnetic field.

Step 3: Determine the magnetic field direction at the location of the right wire.

Using the right-hand rule, your fingers will curl in a clockwise direction around the left wire, which means the magnetic field direction at the location of the right wire is directed into the plane of the page.

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If in a typical thundercloud the bottom of the cloud is negatively charged, what is the sign of the excess charge on the ground?a. negativeb. positivec. the ground is neutral

Answers

The sign of the excess charge on the ground would be positive for a typical thundercloud the bottom of the cloud is negatively charged.

In a typical thundercloud, the bottom of the cloud is negatively charged. When this occurs, it induces an excess charge on the ground. The excess charge on the ground will be opposite in sign to the charge in the cloud. Therefore, the sign of the excess charge on the ground is positive. Every storm cloud has positive power in the upper portion of the cloud, negative power in the lower half, and in many tempests while perhaps not in all there is a concentrated positive charge underneath the super regrettable charge. The droplets and crystals in the air move apart and bump together during the storm. This focuses on making static electrical charges the mists. These clouds have a "plus" and a "minus" end, just like a battery. The cloud's plus, or positive, charges are at the top.

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