what is the capacitive reactance c of a 60.0 μf capacitor placed in an ac circuit driven at a frequency of =123 khz?

Statement is True. When energy is converted from one form to another, a small amount of energy is inevitably lost due to various factors such as friction, heat dissipation, and inefficiencies in the conversion process. This phenomenon is known as energy loss or energy dissipation.

According to the law of conservation of energy, energy cannot be created or destroyed, but it can be transformed from one form to another. However, in practical situations, energy conversions are never 100% efficient, and some energy is always lost in the process. This loss occurs mainly in the form of heat.

The exact calculation of energy loss depends on the specific system and conversion process involved. For example, if electrical energy is converted into mechanical energy using an electric motor, the efficiency of the motor determines the amount of energy lost. Efficiency is usually expressed as a percentage, indicating the ratio of useful output energy to the input energy. The energy lost can then be calculated by subtracting the output energy from the input energy.

In summary, when energy is converted from one form to another, a small amount of energy is inevitably lost due to various factors. This loss is a result of real-world inefficiencies in the conversion process, such as friction and heat dissipation. Therefore, it is important to consider energy losses when designing systems or evaluating the overall efficiency of energy conversion processes.

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In the movie Blade Runner, spotlights are used more frequently than floodlights. Spotlights are often employed to create focused,

directional lighting that highlights specific objects or areas. They provide a concentrated beam of light that can be precisely directed, making them suitable for highlighting key elements or characters in a scene. This use of spotlights helps to create a visually dramatic and atmospheric environment in the film, emphasizing certain details and enhancing the overall cinematic experience. Floodlights, on the other hand, offer a broader and more diffused illumination, often used for general lighting purposes or to cover larger areas. However, in Blade Runner, the predominant use of spotlights adds to the noir-inspired aesthetic and enhances the futuristic ambiance of the film.:

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To answer your question, the amplitude of the wave can be found by dividing the total vertical distance (0.18m) by 2 since amplitude is the distance from the wave's equilibrium position to the highest or lowest point. Thus, the amplitude is 0.09m.

When the amplitude increases to 0.23m, you can determine the factor by which the bug's maximum kinetic energy changes using the relationship between kinetic energy and amplitude. Kinetic energy is directly proportional to the square of the amplitude.

Therefore, to find the factor, you can use the ratio of the squares of the new amplitude (0.23m) and the original amplitude (0.09m).

Factor = (0.23^2) / (0.09^2) ≈ 6.58

So, the bug's maximum kinetic energy increases by a factor of approximately 6.58 when the amplitude increases to 0.23m.

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When the 47-lb package is at the top of the ramp, it has only potential energy since it's at rest. This potential energy is equal to mgh (mass x gravity x height).

Where m is the mass of the package, g is the acceleration due to gravity, and h is the height of the package at the top of the ramp.

As a result, the package's potential energy is [tex]47 x 32.2 x 2 = 3039.4 ft-lbs[/tex].

When the package starts to slide down the ramp, this potential energy is transformed into kinetic energy, which is equal to (1/2) mv^2 (0.5 x mass x velocity^2).Where m is the mass of the package, and v is the velocity of the package at any point in time. If the package is moving at the bottom of the ramp, its kinetic energy is equal to 3039.4 ft-lbs.

Since the ramp is smooth, there is no friction between the ramp and the package, therefore the package's kinetic energy will not be lost as it slides down. The package collides with the spring after descending the ramp, which then compresses the spring, transforming the package's kinetic energy into elastic potential energy stored in the spring.

At this moment, all of the kinetic energy that the package had has been transformed into elastic potential energy of the spring.

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The distance to the galaxy is 50 million light years.

Recessional velocity of the galaxy, v = 1100 km/s

Hubble's constant, H₀ = 22 km/s/mly

Hubble made the amazing discovery that the redshift of galaxies was directly correlated with the distance of the galaxy from the earth. As a result, objects farther from Earth were moving away more quickly. Alternatively said, the universe must be expanding.

According to Hubble's law, the velocity of recession between our galaxy and the other galaxies is directly proportional to the distance between one another.

So, the velocity of recession of the galaxy,

v = H₀ d

Therefore, the distance to the galaxy,

d = v/H₀

d = 1100/22

d = 50 mly

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If Monica took a train 30 miles south to Cincinnati after which she took a different train 18 miles east, then the resultant vector is 35 miles 59 degrees south of east.

A resultant vector is defined as a single vector that produces the same effect as is produced by a number of vectors collectively.

We calculate using  the Pythagorean theorem

c² = a²+ b²

where c =  resultant vector

a and b = individual trip

c² = 30²+ 18²

c² = 1224

c = 35 miles

angle = arctan ( 30 /18)

angle = 59 degrees

In conclusion, Vectors in the opposite direction are subtracted from each other to obtain the resultant vector.

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a) The impedance is 20 ohms.

b) The resistance is 10 ohms.

c) The inductance can be either 462 ohms or 599.8 ohms, depending on whether the positive or negative sign is chosen in the equation for Xl.

(a) The impedance of the circuit can be found using the equation: Z = V/I, where V is the maximum voltage and I is the maximum current. Therefore,

Z = 100 V / 5 A = 20 ohms.

(b) The resistance can be found using the equation: R = Z cos(θ), where θ is the phase angle between the voltage and current. In this case, θ = 60 degrees, so

R = 20 ohms * cos(60 degrees) = 10 ohms.

(c) The reactance of the capacitor can be found using the equation: Xc = 1 / (2πfC), where f is the frequency and C is the capacitance. Plugging in the given values, we get Xc = 1 / (2π * 60 Hz * 5 F) ≈ 530.9 ohms.

The reactance of the inductor can be found using the equation: Xl = 2πfL, where L is the inductance. Since we know the total impedance (Z) and the reactance of the capacitor (Xc), we can use the Pythagorean theorem to find the reactance of the inductor:

Z² = R² + (Xl - Xc)²

(Xl - Xc)² = Z² - R²

Xl - Xc = ±√(Z² - R²)

Xl = Xc ± √(Z² - R²)

Plugging in the given values, we get:

Xl = 530.9 ohms ± √(20² ohms - 10² ohms) ≈ 530.9 ohms ± 68.9 ohms

So the inductance can be either 462 ohms or 599.8 ohms, depending on whether the positive or negative sign is chosen in the equation for Xl.

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The mottled appearance of the Sun's surface, known as granulation, is primarily caused by convective motion in the Sun's outer layer, known as the photosphere.

The granulation pattern arises due to the convection currents that transport heat from the Sun's interior to the surface. As the hot plasma rises, it creates bright regions, or granules, on the surface. These granules represent areas of hotter plasma where the rising material is visible. Simultaneously, the cooler plasma sinks back into the interior, forming darker regions known as intergranular lanes.

The granulation pattern is a manifestation of the complex and dynamic nature of the Sun's outer layers. It is indicative of the convective motions occurring within the Sun, with rising and sinking plasma continuously reshaping the photosphere. The granulation pattern is visible when observing the Sun's surface at high spatial resolution, such as through telescopes equipped with specialized filters or instruments.

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To establish that the universe was very, very large, Edwin Hubble used observations of galaxies and their redshifts.

Edwin Hubble, an American astronomer, made groundbreaking discoveries that revolutionized our understanding of the universe. One of his key contributions was the realization that galaxies exist beyond the boundaries of our Milky Way and that the universe is much larger than previously believed. Hubble observed numerous galaxies and noticed that their light exhibited a redshift. This redshift phenomenon occurs when light waves from distant objects stretch as the universe expands, causing the wavelengths to appear longer, shifting towards the red end of the spectrum. Hubble's observations showed a correlation between the distance of galaxies and the magnitude of their redshifts.

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The height of the image of the person on the film is approximately 1.53 mm.

1/f = 1/do + 1/di

[tex]d_i[/tex] = 47.0 mm / (1/23.4 m - 1/47.0 mm) = 23.3 mm

(image height) / (object height) = (image distance) / (object distance)

(image height) = (object height) x (image distance) / (object distance)

(image height) = 1.535 m x 0.0233 m / 23.4 m ≈ 1.53 mm

Focal length is a fundamental concept in optics and photography that describes the distance between the lens and the image sensor or film when the lens is focused on infinity. It is usually measured in millimeters (mm) and is a critical factor in determining the angle of view and magnification of the image produced by a lens.

A lens with a short focal length, such as a wide-angle lens, has a wider angle of view and can capture a larger area in a single shot. On the other hand, a lens with a longer focal length, such as a telephoto lens, has a narrower angle of view and can magnify distant subjects. Focal length also affects depth of field, which is the range of distances in the scene that appears in sharp focus. A lens with a longer focal length produces shallower depth of field, while a shorter focal length results in deeper depth of field.

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The temperature of the gas of CO2 molecules with an rms speed of 328 m/s is approximately 405 K.

The rms speed of gas molecules is given by the equation v(rms) = sqrt(3kT/m), where k is the Boltzmann constant, T is the temperature in Kelvin, m is the mass of one molecule, and v(rms) is the rms speed. Rearranging this equation to solve for T, we get T = (mv(rms)^2)/(3k). The mass of one CO2 molecule is approximately 44 amu (atomic mass units), or 7.3 x 10^-26 kg. Plugging in this value, as well as the given rms speed of 328 m/s and the Boltzmann constant k = 1.38 x 10^-23 J/K, we get T = (7.3 x 10^-26 kg)(328 m/s)^2/(3*(1.38 x 10^-23 J/K)) = 405 K (rounded to three significant figures).

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The pH of the solution prepared by dissolving 0.15 gram of solid CaO in enough water to make 2.00 L of aqueous Ca(OH)2 is approximately 12.5.

When CaO is dissolved in water, it reacts with water to form Ca(OH)2. This reaction is highly exothermic and releases a significant amount of heat. The resulting Ca(OH)2 is a strong base with a high pH. In this case, the Ca(OH)2 is dissolved in a large volume of water, which further dilutes the solution and lowers the pH slightly. However, since Ca(OH)2 is a strong base, the pH of the resulting solution is still quite high, around 12.5.

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If you increase the distance by a factor of 3, the intensity decreases by a factor of 9. This means that the correct answer is option b.

When you increase the distance by a factor of 3 from a sound source that is radiating equally in all directions, the intensity of the sound decreases. The relationship between distance and intensity is inverse square, meaning that if you double the distance from the source, the intensity decreases by a factor of 4. Therefore, if you increase the distance by a factor of 3, the intensity decreases by a factor of 9.

This means that the correct answer is option b, where the intensity reduces to 1/27th of its original value. It is important to note that this relationship assumes that there are no obstructions or other factors that could affect the sound wave as it travels through the air.

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A square tile that has a 50 cm side is designed to withstand a pressure of 200 N/m². We have to determine if a force of 45 N can be applied to the tile.

The formula for pressure is:

Pressure = Force/Area, where area is in square meters.

To determine the maximum force that can be applied to the tile, we need to first convert the side length of the tile from centimeters to meters.50 cm = 0.5 m

The area of the tile is:

Area = (side length)²Area = (0.5 m)²Area = 0.25 m²

Using the formula for pressure: Pressure = Force/Area200 N/m² = F/0.25 m²

Multiplying both sides by 0.25 m²:50 N = F

Therefore, the maximum force that can be applied to the tile is 50 N, which is greater than the force of 45 N.

Therefore, a force of 45 N can be applied to the tile.

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A physics concept called force defines the impact one thing has on another, causing it to accelerate or change shape. The rate at which an object's momentum changes is measured in newtons (N) by the International System of Units (SI).

a. To find the horizontal and vertical components of the force, we can use trigonometry. Let F be the force you exert on the suitcase, and then the horizontal component of the force (F_h) is given by F_h = F cos(theta), where theta is the angle between the strap and the horizontal. In this case, theta = 45°, so F_h = 30 lb * cos(45°) = 21.2 lb. Similarly, the vertical component of the force (F_v) is given by F_v = F sin(theta), so F_v = 30 lb * sin(45°) = 21.2 lb.

b. If the angle of the strap is 30° instead of 45°, then the horizontal component of the force would be greater. This is because the horizontal component of the force is proportional to the cosine of the angle, and the cosine of 30° is greater than the cosine of 45°. Specifically, F_h = 30 lb * cos(30°) = 25.9 lb.

c. If the angle of the strap is 30° instead of 45°, then the vertical component of the force would be smaller. This is because the vertical component of the force is proportional to the sine of the angle, and the sine of 30° is smaller than the sine of 45°. Specifically, F_v = 30 lb * sin(30°) = 15 lb.

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If one element in a delta-wired three-phase 240 VAC 15 kW electric duct heater is open, the resistance between any two phases would increase by a factor of three.

In a delta-wired three-phase system, each element is connected between two phases, forming a closed loop. If one element is open, the circuit is broken, and the resistance between the two phases connected to the open element increases by a factor of three. This is because, in a delta-wired system, the total resistance is the sum of the individual resistances of each element, and with one element open, the total resistance increases significantly.

In terms of measurements, we could use a multimeter to measure the resistance between any two phases before and after the element is opened. The resistance between the two phases connected to the open element would increase by a factor of three, while the resistance between the other two phases would remain the same. It is important to note that with one element open, the total power output of the heater would be reduced, as the open element would not be contributing to the heating.

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Complete Question:

the resistance on a delta wired three-phase 240 vac 15 kw electric duct heater. if one element is open, what would the measurements be?

The diameter of the coil will be approximately 0.38 meters (or 38 cm). To calculate this, we can use the formula for the magnetic field at the center of a current loop:

[tex]B = (μ₀ * N * I * R²) / (2 * R² + x²)^(3/2)[/tex]

where μ₀ is the permeability of free space, N is the number of turns, I is the current, R is the radius of the loop, and x is the distance from the center of the loop to the point where we want to measure the magnetic field.

We know that N = 2πR / L, where L is the length of the wire, so we can substitute that in and solve for R:

R = L / (2πN) = L / (2π * (L / (2πR))) = R² / L

Substituting that back into the first equation and solving for R, we get:

R = (μ₀ * N * I * L² / 2)^(1/3) / (2π)^(2/3)

Plugging in the given values, we get:

R ≈ 0.38 m

Therefore, the diameter of the coil will be approximately 0.38 meters (or 38 cm).

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Correct option is (d): asking, "On a scale of 1 to 10, how afraid of snakes are you?"

Option (d) asking, "On a scale of 1 to 10, how afraid of snakes are you?" is an operational definition of "fear of snakes" that can be assessed as a structured question.

This question provides a clear and quantifiable measure of the individual's fear level by assigning a numerical value on a scale. It allows participants to express their level of fear in a standardized and structured manner, facilitating easy comparison and analysis of responses.

This approach enables researchers to gather data on the intensity of fear experienced by individuals towards snakes and provides a measurable basis for studying and understanding the phenomenon of fear of snakes.

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The ball would reach a height of approximately 68.88 meters on the moon.

To find the height the ball would reach on the moon, equate the kinetic energy of the ball on Earth to the potential energy it reaches at its maximum height.

On Earth:

Kinetic energy (KE) = Potential energy (PE)

The kinetic energy of the ball can be calculated using the formula:

KE = (1/2) × mv²

Let's assume the mass of the ball is m, and the velocity is v.

On Earth, the kinetic energy of the ball is:

KE_earth = (1/2) × m × (47 m/s)²

The potential energy of the ball at its maximum height on Earth is given by:

PE_earth = m × g × h_earth

where g = acceleration due to gravity on Earth (9.8 m/s²) and h_earth = height reached on Earth (112.59 meters).

Therefore, the equation:

KE_earth = PE_earth

(1/2) × m × (47 m/s)² = m × 9.8 m/s² × 112.59 meters

Now, calculate the height the ball would reach on the moon using the given information that the moon's gravity is 16.7% that of Earth's.

On the Moon:

The acceleration due to gravity on the Moon (g_moon) is 16.7% of Earth's gravity:

g_moon = 0.167 × 9.8 m/s²

The potential energy of the ball at its maximum height on the Moon is given by:

PE_moon = m × g_moon × h_moon

where h_moon = height we need to calculate.

Since the kinetic energy of the ball is conserved (KE_earth = KE_moon), we can set up the equation:

(1/2) × m × (47 m/s)² = m × g_moon × h_moon

Now solve for h_moon:

(1/2) × (47 m/s)² = g_moon × h_moon

h_moon = [(1/2) × (47 m/s)²] / g_moon

Substituting the values:

h_moon = [(1/2) × (47 m/s)²] / (0.167 × 9.8 m/s²)

h_moon = 68.88 meters

Therefore, the ball would reach a height of approximately 68.88 meters on the moon.

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In a series circuit, the total resistance is the sum of all individual resistances. So, if the bottom resistor is changed to 20 ω, the total resistance of the circuit will increase.

The voltage across the resistor can be found using Ohm's law, which states that voltage (V) is equal to the current (I) times the resistance (R), or V=IR.

Assuming a constant current flow through the circuit, the voltage across the 20 ω resistor can be found by using the total resistance of the circuit. Let's say the total resistance is 100 ω. If the current through the circuit is 1A, then using Ohm's law, the voltage across the 20 ω resistor would be V=IR=1A x 20 ω = 20V.

Therefore, if the resistance of the bottom resistor in a series circuit is changed to 20 ω, the voltage across it will depend on the total resistance and current of the circuit. In the example above, the voltage across the 20 ω resistor would be 20V.

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The mechanical advantage of the given pulley system is 3.

Through the use of a tool or system, mechanical advantage is the force multiplier. By definition, a tool or machine will give off a lesser input force over a longer distance for the same amount of work as a larger force over a shorter distance.

In a pulley system, the mechanical advantage is directly proportional to the number of strands connecting to the mass through movable pulleys.

a) Therefore, the mechanical advantage of the given pulley system,

IMA = 3

b) The force or effort applied to pull the weight,

F = Weight/IMA

F = 500/3

F = 166.66 N

c) The efficiency of the machine,

η = IMA/No. of pulleys used

η = 3/3 x 100

η = 100%

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When, a 3.9-mm-diameter wire will carries a 15 A current, and the electric field is 6.9 10⁻² V/m. Then, the wire's resistivity is 6.5 × 10⁻⁸ Ω×m.

We can use Ohm's law to solve for the wire's resistivity, which relates the electric field, current, and wire's dimensions;

ρ = (E × A) / J

where ρ is resistivity, E is electric field, A is cross-sectional area, and J is current density.

First, we need to find the cross-sectional area of the wire;

A = π × (d/2)² = π × (3.9/2)² × 10⁻⁶ m² = 1.19 × 10⁻⁵ m²

where d is the wire's diameter.

Next, we can find the current density;

J = I / A = 15 A / 1.19 × 10⁻⁵ m²

= 1.26 × 10⁶ A/m²

Finally, we can use the equation for resistivity to solve for ρ;

ρ = (E × A) / J = (6.9 × 10⁻² V/m) × (1.19 × 10⁻⁵ m²) / (1.26 × 10⁶ A/m²) = 6.52 × 10⁻⁸ Ω×m

Therefore, the wire's resistivity is 6.5 × 10⁻⁸ Ω×m (in two significant figures).

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In the given image, we have a right triangle with a vertical side of length 5 and a horizontal side of length 12. We are asked to find the length of the hypotenuse.

Using the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides, we can calculate the length of the hypotenuse.

Let's denote the hypotenuse as c. The Pythagorean theorem equation for this triangle is:

c^2 = 5^2 + 12^2

c^2 = 25 + 144

c^2 = 169

Taking the square root of both sides, we find:

c = √169

c = 13

Therefore, the length of the hypotenuse (c) is 13 units.

The angles x and y are both equal to the angle of incidence, which is 30°.

According to the law of reflection, the angle of incidence is equal to the angle of reflection. In this case, the angle of incidence (i) is given as 30°.

Since the angle of incidence is equal to the angle of reflection, the angle of reflection (r) will also be 30°.

To find the values of angles x and y, we can use the fact that the angle of incidence is equal to the angle of reflection. Since x is the angle between the incident ray and the mirror, and y is the angle between the reflected ray and the mirror, both x and y will be equal to the angle of incidence, which is 30° in this case.

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When the air is released from the compressed-air tank into the atmosphere, it expands and occupies a volume of 4.84 m³ at the given temperature and pressure conditions.

To solve this problem, we can use the Ideal Gas Law equation, PV=nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature.

For this problem, we assume the number of moles of air (n) remains constant, so we can rearrange the equation to P₁V₁/T₁ = P₂V₂/T₂.

Initially, the air is in the compressed-air tank with a volume of 0.550 m³ (V₁), temperature of 289 K (T₁), and pressure of 890 kPa (P₁). When released into the atmosphere, the pressure is 101 kPa (P₂) and the temperature is 304 K (T₂). Our goal is to find the final volume (V₂) of the air in these atmospheric conditions.

Plugging the values into the equation,

we have (890 kPa × 0.550 m³) / 289 K = (101 kPa × V2) / 304 K.

After performing the calculations and solving for V₂, we find that the final volume of the air in the atmospheric conditions is approximately 4.84 m³.

Thus, when the air is released from the compressed-air tank into the atmosphere, it expands and occupies a volume of 4.84 m³ at the given temperature and pressure conditions.

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The answer to the question is dependent on several factors, such as the angle of the ladder, the coefficient of static friction between the ladder and the ground, and the weight of the ladder. Without this information, it is not possible to provide a specific answer to the question.

However, in general, the force of static friction on the ladder from the ground can be calculated using the equation F_friction = μ_s*N, where μ_s is the coefficient of static friction and N is the normal force acting on the ladder. The normal force is the force perpendicular to the surface the ladder is resting on and is equal in magnitude to the weight of the ladder. Therefore, the force of static friction can be increased by increasing the coefficient of static friction or the weight of the ladder, or by decreasing the angle between the ladder and the ground. It is important to note that the force of static friction must be greater than or equal to the weight of the ladder in order for the ladder to remain at rest.

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The  exhaust velocity of the hot gases is approximately 1,800 m/s.

To find the exhaust velocity of the hot gases, we can use the ideal rocket equation, which relates the exhaust velocity (Ve) to the thrust of the rocket (F) and the rate at which it burns fuel (m_dot):

Ve = F / m_dot

In this case, the thrust of the rocket is 15 kN (kilonewtons), or 15,000 N, and it burns 250 kg of fuel in 30 seconds. To find the rate at which it burns fuel (m_dot), we can divide the total mass of fuel burned by the time it takes to burn it:

m_dot = 250 kg / 30 s = 8.33 kg/s

Now we can substitute these values into the ideal rocket equation:

Ve = (15,000 N) / (8.33 kg/s) = 1,800 m/s

Therefore, the exhaust velocity of the hot gases is approximately 1,800 m/s.

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According to the rule for current in parallel circuits, the current through each branch of the circuit is determined by the resistance of that branch and the total voltage applied to the circuit. In a series circuit, there is only one path for the current to flow, so the same current flows through every element.

In a parallel circuit, on the other hand, there are multiple paths for the current to flow, and the current through each branch is determined by the resistance of that branch and the total voltage applied to the circuit. Because the voltage is the same across each branch of a parallel circuit, the current through each branch will depend only on the resistance of that branch. Therefore, the current through each branch can be different in a parallel circuit. In contrast, because there is only one path for the current to flow in a series circuit, the same current flows through every element.

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Answer:

portal of entry link

Explanation:

This what I got hope that helps

The statement that is true about quasars is: "All quasars are active galactic nuclei, but not all active galactic nuclei are quasars."

A quasar is a type of active galactic nucleus (AGN) that is highly luminous and characterized by a compact region at the center of a massive galaxy that emits enormous amounts of energy. Quasars are not discovered by identifying pairs of enormous radio-emitting lobes of gas, which is a characteristic of radio galaxies.

Quasars do not appear bright because they gravitationally lens background light; instead, their high luminosity is due to the accretion of matter onto a supermassive black hole at the center of the galaxy. Quasars do not generate their tremendous luminosities via nuclear fusion inside their event horizons, but rather from the frictional heating and gravitational energy released by the infalling matter. Finally, quasars can fluctuate in brightness but are not comparable in size to their host galaxies, as they are typically much smaller.

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When the block is inverted so that the steel is under water, the amount of the block submerged will remain the same. The portion of the block that was previously above the water level will now be submerged, while the portion that was submerged will now be above the water level.

In summary, when the block is inverted, the amount of the block submerged will increase and the water level in the tub will decrease. This is because the weight of the block acting on the submerged portion of the block is greater than the weight of the submerged portion of the block, causing the block to be lifted out of the water, and the volume of the submerged portion of the block is greater than the volume of the block, causing the water level to decrease.  

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Full Question ;

O A small piece of steel is tied to a block of wood. When the wood is placed in a tub of water with the steel on top, half of the block is submerged. Now the block is inverted so that the steel is under water. (i) Does the amount of the block submerged (a) increase, (b) decrease, or (c) remain the same? (ii) What happens to the water level in the tub when the block is inverted? (a) It rises. (b) It falls. (c) It remains the same.