a 62.0-kg woman runs up a 4.28-m high stairway in a time of 4.20 s. what average power did she supply?

The work required to bring the hollow cylinder from rest to an angular velocity of 0.305 m/s without the hoop slipping is 305.76 joules

To calculate the work required to bring the hollow cylinder from rest to an angular velocity of 0.305 m/s, we need to use the formula,

work = (1/2)I[tex]w^{2}[/tex]

where I is the moment of inertia of the cylinder, and ω is the angular velocity.

From the problem statement, we know that the moment of inertia of the cylinder is [tex]mR^2[/tex], where m is the mass of the cylinder and R is its radius. Plugging in the given values, we get

I = [tex](73 kg)(3.05 m)^2[/tex] = 6665.35 [tex]kg m^2[/tex]

Next, we plug in the given angular velocity,

ω = 0.305 m/s

Now we can calculate the work,

work = [tex](1/2)(6665.35 kg m^2)(0.305 m/s)^2[/tex] = 305.76 joules

Therefore, the work required to bring the hollow cylinder from rest to an angular velocity of 0.305 m/s is 305.76 joules.

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James Prescott Joule was the scientist who performed the first experiment demonstrating the equivalence of mechanical energy and heat.

The first experiment that demonstrated the equivalence of mechanical energy and heat was performed by James Prescott Joule in the mid-19th century. Joule's experiment involved measuring the increase in temperature of water as it was stirred by paddles driven by falling weights.

He showed that the amount of mechanical work done by the falling weights was equivalent to the amount of heat generated in the water. This work was instrumental in establishing the principle of conservation of energy, which states that energy cannot be created or destroyed, only converted from one form to another.

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If a bar magnet is held stationary next to a solenoid a current is generated in the magnet

A magnet is a material that can produce a magnetic field and attract certain materials, such as iron, steel, nickel and cobalt. Magnets have two poles, north and south, that attract opposite poles and repel like poles. Magnets have been used for centuries for a variety of purposes, such as navigation and medicine. Magnets can be natural, such as lodestones, or manufactured, such as ceramic magnets and rare earth magnets. Magnets come in a variety of shapes and sizes, including bar magnets, horseshoe magnets, disc magnets and cylinder magnets. Permanent magnets are made of materials that retain their magnetism, while temporary magnets can only be magnetized while in an external magnetic field. Magnets can be used to store data on floppy disks, hard drives and credit cards, and to generate electricity in generators and motors.

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We can use the conservation of momentum principle, which states that the total momentum of a system before a collision is equal to the total momentum of the system after the collision. In this case, the system consists of the child and the skateboard.

Before the collision, the momentum of the child is:

p_ child = m_ child * v_ child
p_ child = 30 kg * 4 m/s
p_ child = 120 kg*m/s

Since the skateboard is stationary, its momentum is zero:
p _skateboard = 0 kg*m/s

The total momentum before the collision is:
p_ before = p_ child + p_ skateboard
p_ before = 120 kg*m/s + 0 kg*m/s
p_ before = 120 kg*m/s

After the collision, the child and the skateboard move together as one system. Let's assume their final velocity is v_ final. The total momentum after the collision is:
p_ after = (m_ child + m_ skateboard) * v_ final

Substituting the values we know:
p_ after = (30 kg + 10 kg) * v_ final
p_ after = 40 kg * v_ final

According to the conservation of momentum principle, p_ before = p_ after, so we can set these two equations equal to each other:

p_ before = p_ after
120 kg*m/s = 40 kg * v_ final

Solving for v_ final:

v_ final = 3 m/s

Therefore, the speed of the child and the skateboard after the collision is approximately 3 m/s.

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The direction of sound travel compares to the shapes of the sound waves.

The direction of sound travel is determined by the propagation of the sound waves through a medium, such as air or water. Sound waves are longitudinal waves, which means that the particles in the medium vibrate parallel to the direction of the wave's travel. However, the direction of sound travel is perpendicular to this vibration, meaning that it travels in a straight line away from the source of the sound. This is why we can hear sound from different directions, even though the waves themselves are moving in a specific direction

In other words, when the sound waves move, they cause the particles to compress and rarefy (move closer together and further apart) in the same direction as the wave's movement.

To summarize, the direction of sound travel is parallel to the shapes of the sound waves, since sound waves are longitudinal waves that cause particles in the medium to vibrate along the same direction as the wave's movement.

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Answer:

P = 2 π (L / g)^1/2     describes period of a simple pendulum

The period of the pendulum does not depend on the mass (weight)

Pendulum b has the same period (frequency) as pendulum a

The efficiency of the engine is 83.4% assuming  that the

heat capacity is 21 J/K and the universal gas constant is 0.08206 L · atm/mol/K =8.314 J/mol/K.

Efficiency is described as the often measurable ability to avoid wasting materials, energy, efforts, money, and time while performing a task.

The efficiency of the engine is given by:

E = W/Q

where;

W = the work done in the four steps,

Q = the energy input

Since there at four steps in a cycle:

E = w1+ w2 +w3+ w4/ q1+ q2+q3+q4

We calculate that the work done in the first step (isothermal expansion)

n= 1 mole, T1 = 402 K, V2 = 41.2 L, V1 = 18.5 L

We also solve for Steps 2 and 4 are constant volume processes,

We also calculate  work done in the third step (isothermal expansion) is

where;

n = 1 mol, T3 = 273 K, V4 = 41.2 L, V3 = 18.5 L

We notice that Heat enters the system only during steps (1) and (4).

The internal energy of the gas increases in step 4 but no work is done, while the internal energy is constant change in step 1 but work is done by the gas.

Cv =21 J/K, T3 = 273 K, T4 = 402 K

We Solve  for efficiency, ɛ:

ɛ = 2676.01  +0 +0 + 1879.29/ 2676.01  +0 +0 + 2709 = 83.4%.

Therefore, the efficiency of the engine is 83.4%.

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The torque on the particle about the point with coordinates (-1.00 m, -2.00 m, 5.00 m) is:

τ = (-5.00 Nm)i^ + (15.0 Nm)j^ + (13.0 Nm)k^

The torque τ about a point is given by the cross product of the vector from the point to the particle and the force acting on the particle:

τ = r x F

where r is the vector from the point to the particle, and x denotes the cross product.

We can find the vector r by subtracting the coordinates of the point from the coordinates of the particle:

r = r_particle - r_point

r = (2.00 m, 3.00 m, 4.00 m) - (-1.00 m, -2.00 m, 5.00 m)

r = (3.00 m, 5.00 m, -1.00 m)

Now we can calculate the torque by taking the cross product of r and F:

τ = r x F

τ = (3.00 m, 5.00 m, -1.00 m) x (5.0 N)i^ + (-1.00 N)k^

τ = (5.00 N)(-1.00 m)i^ + (15.0 Nm)j^ + (13.0 Nm)k^

Therefore, the torque on the particle about the point with coordinates (-1.00 m, -2.00 m, 5.00 m) is:

τ = (-5.00 Nm)i^ + (15.0 Nm)j^ + (13.0 Nm)k^

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Conservative forces are independent of path, whereas nonconservative forces are path dependent.

Nonconservative forces are forces that do work on an object that depends on the path taken by the object, rather than just its initial and final positions. Examples of nonconservative forces include friction, air resistance, and tension in a rope that is being stretched. When nonconservative forces are present, the equation of deltaE= delta U + delta K still holds, but the change in energy (deltaE) must take into account the work done by the nonconservative forces.
Conservative forces, on the other hand, are forces that do not depend on the path taken by the object, only on its initial and final positions. Examples of conservative forces include gravity and electrostatic forces.

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The magnitude of the force on the charge is 240 N.

To calculate the magnitude of the force on the charge in a uniform electric field, you can use the following formula:

F = q * E

where F is the force, q is the charge, and E is the electric field magnitude.

In this case, the uniform electric field has a magnitude of 40 N/C, and the charge is -6 C. Plug these values into the formula:

F = (-6 C) * (40 N/C)

F = -240 N

Therefore, the magnitude of the force on the charge is 240 N. Note that the negative sign indicates that the force is directed downward.

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During a thunderstorm, it is safer to be in your car than standing under a tree on a golf course.

This is because lightning can strike tall objects such as trees and golf clubs, making you more susceptible to getting struck.

Being in a car, on the other hand, provides a safe shelter as long as you avoid touching any metal parts inside the vehicle. Remember, it is always best to seek shelter indoors during a thunderstorm.

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Answer: estimate could be anywhere from 15 to 30 minutes, or longer.

Explanation: If your vehicle has incurred significant mechanical and exterior damage, an estimate could be anywhere from 15 to 30 minutes, or longer. When the damage is minimal and mechanical issues don't exist, an estimate usually takes 15-20 minutes.

The dart will oscillate with an angular frequency of approximately 56.47 rad/s.

In physics, angular frequency "ω" is a scalar measure of rotation rate. It refers to the angular displacement per unit time or the rate of change of the phase of a sinusoidal waveform, or as the rate of change of the argument of the sine function.

To find the angular frequency with which the dart oscillates, we can use the spring constant, the mass of the dart, and the equation for angular frequency.

Step 1: Identify the spring constant (k) and the mass of the dart (m).
k = 220 N/m
m = 0.069 kg

Step 2: Use the equation for angular frequency (ω) in a spring-mass system.
ω = √(k/m)

Step 3: Plug the values of k and m into the equation and solve for ω.
ω = √(220 N/m / 0.069 kg)

Step 4: Calculate the angular frequency.
ω ≈ 56.47 rad/s

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Stars that are visible in the local sky on any clear night of the year, at any time of the night, are called circumpolar.

As one moves farther north or south from the equator, the position of the celestial equator in the sky shifts, and the angle between the celestial equator and the horizon changes. This means that some stars that are located near the celestial poles can remain above the horizon at all times, even as the Earth rotates.

For an observer located at a latitude of 90 degrees (the North or South Pole), all stars are circumpolar, and none of them will set below the horizon. For an observer located at a latitude of 30 degrees north, the stars that are within 30 degrees of the north celestial pole are circumpolar, while those within 30 degrees of the south celestial pole will never be visible from that location.

In the Northern Hemisphere, some well-known circumpolar stars include Polaris (also known as the North Star), which is located very close to the north celestial pole, and the stars of the Big Dipper, which circle around Polaris in a counterclockwise direction.

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The three cardinal wave rules are:
Superposition

Reflection

Refraction

1. Superposition Principle: When two or more waves meet at a point, the resultant wave's displacement is the algebraic sum of the individual wave displacements at that point.

2. Reflection: When a wave encounters a boundary or obstacle, it bounces back, changing direction while maintaining its speed, wavelength, and frequency.

3. Refraction: When a wave passes from one medium to another, its speed changes, causing the wave to change direction. This bending of the wave is known as refraction, and it's governed by Snell's Law.

These rules help to describe the behavior of waves in various situations and are fundamental to understanding wave interactions and propagation.

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The electron orbits would be circular with a radius of

[tex]r = mv/( qB)[/tex],

where v is the initial speed of the electrons and q is the charge of the electron, and the direction of the electron orbits would be perpendicular to the magnetic field and in the plane perpendicular to the initial velocity of the electrons.

The motion of a charged particle in a magnetic field is given by the Lorentz force equation:

[tex]F = qvB sin(θ)[/tex]

where F is the force on the charged particle, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field, and θ is the angle between the velocity of the particle and the magnetic field.

Since the force is perpendicular to the velocity, it causes the electrons to move in a circular path around the magnetic field. The radius of this circular path is given by:

[tex]r = mv/(qB)[/tex]

where m is the mass of the electron.

Therefore, the electron orbits would be circular with a radius of

[tex]r = mv/(qB),[/tex] where v is the initial speed of the electrons and q is the charge of the electron. The direction of the electron orbits would be perpendicular to the magnetic field and in the plane perpendicular to the initial velocity of the electrons.

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The acceleration of the second disk would be half of the acceleration of the first disk.

This is because the torque applied to both disks is the same, but the moment of inertia of the second disk is four times greater than that of the first disk (since the moment of inertia is proportional to the radius squared). Therefore, the second disk will have a larger moment of inertia, which means it will be harder to accelerate.

However, the same torque will produce the same angular acceleration in both disks. Since the second disk has twice the radius, its linear acceleration will be half that of the first disk.

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As the mass oscillates up and down,

1. Kinetic energy (KE) will change periodically. It will be at its maximum when the mass is at the midpoint of its oscillation, and at its minimum (zero) when the mass reaches the highest and lowest points in its oscillation.

2. Elastic potential energy (EPE) will also change periodically. It will be at its maximum when the mass is at the highest and lowest points in its oscillation (where the spring is most compressed or stretched), and at its minimum (zero) when the mass is at the midpoint of its oscillation.

3. Mechanical energy (ME), which is the sum of kinetic energy and elastic potential energy, will remain constant throughout the oscillation, as long as there are no external forces (e.g., friction) causing energy loss. This is because, as the mass oscillates, the energy is transferred between kinetic energy and elastic potential energy, but the total energy remains the same.

As the mass oscillates up and down, the kinetic energy will continuously alternate between maximum and zero values. At the highest points of the oscillation, the kinetic energy will be zero as the mass momentarily stops moving. At the lowest points of the oscillation, the kinetic energy will be at its maximum as the mass is moving at its highest speed.

The elastic potential energy will also change in a similar fashion, oscillating between maximum and zero values. At the highest points of the oscillation, the elastic potential energy will be at its maximum as the spring is stretched to its greatest extent. At the lowest points of the oscillation, the elastic potential energy will be at its minimum as the spring is relaxed.

The mechanical energy, which is the sum of the kinetic and potential energies, will remain constant as the mass oscillates. This is because energy is conserved in an elastic system, meaning that the total mechanical energy will remain constant even as the kinetic and potential energies alternate.

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A force is applied for 23 ms to a 0.61 kg ball initially moving at 27 m/s in the positive x direction. The impulse has a magnitude of 45.9 N s, and the average magnitude of the force on the ball is approximately 2.00 kN.

We can use the impulse-momentum theorem to find the change in momentum of the ball during the time the force is applied

J = Δp = [tex]p_f[/tex] - [tex]p_i[/tex]

where J is the impulse, [tex]p_f[/tex] is the final momentum, and [tex]p_i[/tex] is the initial momentum. We know that the ball has an initial momentum of

[tex]p_i[/tex] = m*[tex]v_i[/tex] = 0.61 kg * 27 m/s = 16.47 kg m/s

We also know that the impulse has a magnitude of

J = Δp = [tex]p_f[/tex] - [tex]p_i[/tex] = 45.9 N s

Therefore, the final momentum of the ball is

[tex]p_f[/tex] = [tex]p_i[/tex] + J = 16.47 kg m/s - 45.9 N s = -29.43 kg m/s

The negative sign indicates that the ball is moving in the negative direction of the x axis after the force is applied.

We can find the average magnitude of the force by using the formula

[tex]F_{avg}[/tex] = J / Δt

where Δt is the time interval during which the impulse is applied. We know that the impulse has a magnitude of

J = 45.9 N s

And we are told that the applied force is

Δt = 23 ms = 0.023 s

Therefore, the average magnitude of the force is

[tex]F_{avg}[/tex] = J / Δt = 45.9 N s / 0.023 s ≈ 1996.96 N ≈ 2.00 kN

So the average magnitude of the force on the ball is approximately 2.00 kN.

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a. To find the period of oscillation of a large simple pendulum with a length of 2m and a mass of 3kg, we can use the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. Plugging in the values, we get T = 2π√(2/9.81) ≈ 2.02 seconds.

b. If the length of the pendulum is doubled to 4m, we can use the same formula to find the new period. Plugging in the new value for L, we get T = 2π√(4/9.81) ≈ 4.04 seconds. So doubling the length of the pendulum results in a doubling of the period.

c. If the length of the pendulum is shortened back to 2m, but the mass is doubled to 6kg, we can again use the same formula to find the new period. Plugging in the new values for L and the mass, we get T = 2π√(2/9.81) ≈ 1.43 seconds. So doubling the mass while keeping the length constant results in a shorter period.

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When a charged rod is brought near bits of uncharged paper, the paper will become polarized. This means that the positive charges in the paper will be attracted to the negatively charged rod, and the negative charges in the paper will be repelled by the rod.

                                   This will cause the bits of paper to move towards the rod and potentially stick to it, even though they are themselves uncharged. This is because the polarized charges in the paper are attracted to the opposite charges on the rod.

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The distance between two consecutive nodes in a wave is half the wavelength.

The distance between nodes is related to the wavelength in the following way:
The distance between two consecutive nodes in a wave is half the wavelength. This is because a node is a point where the wave has zero amplitude, and consecutive nodes are separated by one complete cycle of the wave, which is equal to half of the wavelength.

1. Identify the nodes in a wave, which are the points of zero amplitude.
2. Measure the distance between two consecutive nodes.
3. Multiply that distance by 2 to find the wavelength of the wave.

In summary, the distance between nodes is half the wavelength of the wave.

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When the apple falls from the limb, its potential energy is converted into kinetic energy as it gains speed while falling towards the ground. Just before the apple hits the ground, its kinetic energy is at its maximum, while its potential energy is at its minimum.

The potential energy of an apple hanging from a limb is converted into other forms of energy as it falls and eventually hits the ground.

Just before the apple hits the ground, its potential energy has been converted mainly into kinetic energy due to its motion. This happens because as the apple falls, it gains speed, and the potential energy is gradually transformed into kinetic energy.

When the apple hits the ground, the kinetic energy is then transferred into other forms of energy, such as sound energy, heat energy, and deformation (or internal) energy within the apple as it impacts the ground and becomes slightly compressed or damaged.

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To calculate the kinetic energy of the ball we will use the equation KE = 1/2 * m * v^2.

The kinetic energy (KE) of an object is defined as the energy that it possesses due to its motion. It is given by the formula:

KE = 1/2 * m * v^2

where:

m is the mass of the object

v is the velocity of the object

To calculate the kinetic energy of a ball that is dropped and whose velocity is known at a certain time, we would use this formula by plugging in the values of the mass and velocity. For example, suppose that the mass of the ball is 0.2 kg and its velocity is 5 m/s at a certain time. Then the kinetic energy of the ball at that time would be:

KE = 1/2 * m * v^2

= 1/2 * (0.2 kg) * (5 m/s)^2

= 2.5 J

So, the kinetic energy of the ball at that time would be 2.5 joules.

Note that the kinetic energy of an object increases with its mass and the square of its velocity. This means that an object with a larger mass or a higher velocity will have greater kinetic energy than an object with a smaller mass or a lower velocity. Additionally, the kinetic energy of an object can be converted into other forms of energy, such as potential energy or heat, when it interacts with other objects or systems.

Therefore, the " KE = 1/2 * m * v^2 " this equation is used for calculating kinetic energy.

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Power supplied to the lamp in Watts = 3.2 W,

The voltage that produces this current  = 578 V,

The resistance of the circuit in Ohms:  = 0.074 Ω

For question 31:

Using the formula P = I^2 * R, we can find the power supplied to the lamp:

[tex]P = (0.8 A)^2 * 5 \Omega = 3.2 W[/tex]

For question 32:

Using Ohm's Law, we can find the voltage:

V = I * R,

[tex]V = 7 A * (578 W / 7 A) = 578 V[/tex]

For question 33:

Using the formula P = V^2 / R, we can find the resistance of the circuit:

R = V^2 / P, where R is resistance in ohms, V is voltage in volts, and P is power in watts.

[tex]R = (2.1 V)^2 / 59 W = 0.074 \Omega[/tex]

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A sawtooth wave is commonly used in subtractive synthesis as the source for a bowed string instrument because it closely resembles the harmonic content of a bowed string.

The sawtooth wave contains a rich set of overtones that are well-suited for creating the complex, expressive tones of a bowed string instrument. Additionally, the sawtooth wave's sharp rise and fall mimics the attack and decay of a bowed string, allowing for more realistic sound synthesis.

Finally, by applying a low-pass filter to the sawtooth wave, the higher harmonics can be gradually removed, creating the characteristic "mellow" sound of a bowed string instrument.

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A uniform sphere of radius r and mass m rotates freely about a horizontal axis that is tangent to an equatorial plane of the sphere. The moment of inertia (I) of the sphere about this axis can be calculated using the parallel axis theorem.

Step 1: Find the moment of inertia for a uniform sphere about its center of mass. For a sphere, this value is given by the equation:

I_center = (2/5) * m * r^2

Step 2: Apply the parallel axis theorem. The parallel axis theorem states that the moment of inertia (I) about an axis parallel to and a distance d away from the axis through the center of mass is:

I = I_center + m * d^2

Step 3: In our case, the distance d is equal to the radius r, since the horizontal axis is tangent to the equatorial plane of the sphere. Plug this value and I_center into the parallel axis theorem equation:

I = (2/5) * m * r^2 + m * r^2

Step 4: Simplify the equation:

I = m * r^2 * ((2/5) + 1)

I = m * r^2 * (7/5)

So, the moment of inertia of the uniform sphere about the horizontal axis tangent to the equatorial plane is:

I = (7/5) * m * r^2

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When two objects are traveling in a circular orbit. The centripetal acceleration of object A is twice that of object B.

Centripetal acceleration is the acceleration of the body that travels in a circular motion. Any object that moves in the circular path and its vector is towards to the center is called as Centripetal acceleration. It is obtained by the ratio of the velocity square and the radius of the circle.

From the givens,

Object A moves with the velocity twice that of Object B and the diameter of  the circle moves by object A is twice that of the diameter of circle moves by object B. Acceleration (a) = v² / r.

Object B's acceleration, a =  v² / 2r ( diameter d = 2r)

Object A's acceleration, a = (2v)² / (2r)

                                           =  4/2 (v²/ 2r)

                                           = 2 (v²/ 2r).

Thus, the object A's acceleration is twice that of acceleration of object B.

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To determine the temperature of the kiln, we can use the concept of heat transfer, where the heat gained by water equals the heat lost by the copper block.

First, we find the heat gained by water:

Q = mcΔT

where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Q_water = m_water * c_water * ΔT_water

Where:
m_water = 300 g
c_water = 1.00 cal/g·°C
ΔT_water = (35°C - 15°C) = 20°C

Q_water = 300 g * 1.00 cal/g·°C * 20°C = 6000 cal

Now, we find the heat lost by the copper block:
Q_copper = m_copper * c_copper * ΔT_copper

Where:
m_copper = 120 g
c_copper = 0.10 cal/g·°C
ΔT_copper = (T_kiln - T_final_copper)

Since heat gained by water equals heat lost by copper:
Q_water = Q_copper

6000 cal = 120 g * 0.10 cal/g·°C * (T_kiln - 35°C)
6000 cal = 12 cal/°C * (T_kiln - 35°C)

Now, solve for T_kiln:
500 = T_kiln - 35°C
T_kiln = 535°C

The temperature of the kiln was 535°C (option d).

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