The data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1000 K and 300 K, respectively. For each case, determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
a) QH = 600 kJ, Wcycle = 300 kJ, QC = 300 kJ
b) QH = 400 kJ, Wcycle = 280 kJ, QC = 120 kJ
c) QH = 700 kJ, Wcycle = 300 kJ, QC = 500 kJ
d) QH = 800 kJ, Wcycle = 600 kJ, QC = 200 kJ

Answer:

The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.

Explanation:

Answer:

Im not really sure lemme ask my friend he knows about this subject and when he gives me answer ill edit this

Explanation:

Answer:

i dont even know

Explanation:

I just dont know

\(\text{Kinetic energy,}\\\\E_k = \dfrac 12 mv^2\\\\~~~~~=\dfrac 12 \times 0.056 \times 44^2\\\\~~~~~=54.208~J\)

The magnitude of the acceleration of the ball  applied at the bottom of the circle is expressed as FTension-FGravity/M.

You can write it down as FC = mv2 /r.

The centripetal force in a moving object attached to a string is equal to the product of the object's mass (mg) and the string tension (T).

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Answer:

Explanation:

Formula

Force = G * m1 * m2 / r^2

Givens

Solution

1.67 * 10^-9 = 6.67 * 10^-11 * 110 * m2/(11)^2        Divide both sides by 10^-9

1.67 * 10^-9/10^-9 = 6.67*10^-11/(10^-9 *110*m2/121

1.67 = 6.67 * 110 * 10^-2 * m2 / 121                      Multiply both sides by 121

1.67 * 121  = 6.67 * 10 ^-2 * 110 * m2 * 121/121      Combine

202.07 = 7.337 * m2                                           Divide both sides by 7.337

202.07/7.337  =  m2

m2 = 27.54 kg

The average distance travelled by the muons is 5.50 km in the Earth's frame of reference.

Muon is a subatomic particle that is a fundamental constituent of matter. It is classified as a lepton, along with the electron, tau, and three neutrinos. A muon's rest mass is 105.65837 MeV/c², which is around 207 times greater than the electron's rest mass. A muon's rest lifetime is 2.2 microseconds.

Find the average distance covered by the muons if their speed relative to Earth is 0.845c. The muon's lifetime can be used to determine the average distance it travels if its speed is constant over that time. The distance can be calculated using the following formula:

Distance = Speed × Time

A muon's lifetime of 2.2 microseconds and a relative velocity of 0.845c are given. We can use the above formula to determine the average distance covered by a muon in this situation.

Distance = Speed × Time= 0.845c × 2.2 µs= 4.97 × 10⁻⁴ km or 497 meters.

Since the muons are travelling towards Earth from a height of 5.00 km, we can add the height of their point of creation to the distance they travelled to determine the average distance they travelled from creation to the Earth's surface.

Average distance travelled by muon = Distance + Height= 497 m + 5.00 km= 5.50 km (to 3 significant figures).

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Answer:

I believe the answer is D: Conductor.

The hare has the greater instantaneous speed since he can go faster.

In a race, both the turtle and the hare travel the same distance. The tortoise travels gradually and finishes the race before the hare, whereas the hare travels very quickly for brief periods of time but frequently stops.

Since he can move more quickly, the hare has a faster instantaneous speed. Since the turtle won, he moves at a faster average speed.

An object's instantaneous speed is a measure of how quickly it is travelling right now. The speed and direction of an object's motion are both included in its instantaneous velocity, which is a vector quantity.

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Answer:

the velocity is 1.67 m/s

Explanation:

The computation of the velocity of the arrow/ice combination after the collison is shown below:

Here the law of conservation is applied

\(p_i = p_f\\\\m_av_a+m_b+v_b= (m_a + m_b)v\)

Here

\(m_a = 2.0 kg\\\\v_a = 10m/s\\\\m_b = 10 kg\\\\v_b = 0\)

Now

\(v = \frac{m_av_a+ m_bv_b}{m_a + m_b} \\\\= \frac{(2.0kg) (10 m/s) + 0}{2.0kg + 10.0kg}\)

= 1.67 m/s

hence, the velocity is 1.67 m/s

Answer:

3, 4, 1, 2

Explanation:

The magnitude of the second force is 1710 N.

Force is the product of mass and acceleration.

To calculate the magnitude of the second force, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

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The length of the copper wire must be approximately 44.2 meters for it to have a resistance of 0.128 ohms, assuming a 14-gauge wire with a diameter of 1.63 mm and using the resistivity of copper.

To find the length of the wire, we can use Ohm's Law, which states that the resistance (R) is equal to the product of the resistivity (ρ), the length (L), and the cross-sectional area (A) of the wire, divided by the diameter (d) of the wire squared.

The formula can be written as:

R = ρ * (L / A)

Resistance (R) = 0.128 ohms

Resistivity of copper (ρ) = (1.68 × 10^-8) ohm-meter (at 20°C)

Diameter (d) = 1.63 mm = 0.00163 meters (converted from millimeters to meters)

We need to find the length (L) of the wire.

To calculate the cross-sectional area (A) of the wire, we can use the formula for the area of a circle:

A = π * (d/2)^2

Plugging in the values, we have:

A = 3.14159 * (0.00163 / 2)^2

A  ≈ 2.08 x 10^-6 square meters

Rearranging Ohm's Law to solve for the length (L), we get:

L = (R * A) / ρ

Substituting the given values:

L = (0.128 * 2.08 x 10^-6) / (1.68 x 10^-8)

L  ≈ 44.2 meters

The length of the copper wire must be approximately 44.2 meters for it to have a resistance of 0.128 ohms, assuming a 14-gauge wire with a diameter of 1.63 mm and using the resistivity of copper.

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If the distance separating Objects 1 and 2 is doubled, then the new gravitational force will be 72.0 units.

The force of attraction that operates on all matter is known as gravity, or gravitational force in mechanics. It is by far the weakest known force in nature and plays no role in determining the internal properties of everyday matter. It controls the orbits of objects in places, the structure, and the evolution of stars, galaxies, and the universe as a whole.

On Earth, every object has a weight, or downward gravitational force, proportional to its mass, and the mass of the Earth acts on them. The acceleration that gravity causes in a falling item serves as a gauge of its strength. The acceleration of gravity at the Earth's surface is around 9.8 meters per second (32 feet) per second. For each second her body is in free fall, its velocity increases by about 9.8 meters per second. On the moon, the acceleration of a free-falling object is approximately 1.6 meters per second.

The gravitational force among two objects can be written as:

F = G×m₁×m₂/r²

Where G is the gravitational constant, m₁ and m₂ are the masses of both objects and r is the distance between them.

Now, we have that F = 18 .0 units.

Let's change r to 2r, the new force is:

F' = G×m₁×m₂/(2r)²

F' = G×m₁×m₂/4r²

F'/F = (G×m₁×m₂/4r²)/G×m₁×m₂/r²

F' = 4 F

F' = 4 × 18.0

F' = 72 units

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The capacitance of an air-filled capacitors does not change is if electric field in between plates is made weaker by discharging the plates:

The external unit's malfunctioning due to a defective capacitor impedes the cooling procedure as a whole. Second, the system must work harder to complete its task due to poor voltage distribution to outer unit components. A defective capacitor frequently causes harm to additional components.

In a simple parallel-plate capacitor, applying a voltage between 2 conducting plates creates an equal electric field between both the plates. The electrical field intensity in a capacitor is inversely related to the distance between the plates but directly proportional to the applied voltage.

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Visible radiations are the electromagnetic radiation wavelength which is efficiently observed using ground based telescopes.

The electromagnetic waves we can for the most part see on the ground comprise of apparent light, which is challenging for the air to retain, close infrared beams, and a few electromagnetic waves. These frequency ranges are called air windows.

Noticeable light is a sort of electromagnetic radiation, which is sent in waves or particles at various frequencies and frequencies. This wide scope of frequencies is known as the electromagnetic range. That range is commonly separated into seven districts arranged by diminishing frequency and expanding energy and frequency.

A commonplace natural eye will answer frequencies from around 380 to around 750 nanometers. As far as recurrence, this compares to a band nearby 400-790 terahertz. These limits are not pointedly characterized and may differ per person. Under ideal circumstances these restrictions of human insight can stretch out to 310 nm (bright) and 1100 nm (close to infrared).

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Given the following information:Mass of the system, m = 20 kg.Damping coefficient, b = 6 Ns/m.Force, F = 90 N.Frequency of applied force, f = ?Applied force angular frequency, w = 6 rad/s.Forced vibration equation:F(t) = F0 sin(wt)where F0 = 90 N and w = 6 rad/s.Under the action of the force F, the mass m will oscillate.The equation of motion for the mass-spring-damper system is given by:$$\mathrm{m\frac{d^{2}x}{dt^{2}}} + \mathrm{b\frac{dx}{dt}} + \mathrm{kx = F_{0}sin(\omega t)}$$where k is the spring constant.x(0) = 0 and x'(0) = 0.As we have the damping coefficient (b), we can calculate the damping ratio (ζ) and natural frequency (ωn) of the system.Damping ratio:$$\mathrm{\zeta = \frac{b}{2\sqrt{km}}}$$where k is the spring constant and m is the mass of the system.Natural frequency:$$\mathrm{\omega_{n} = \sqrt{\frac{k}{m}}}$$where k is the spring constant and m is the mass of the system.Resonant frequency:$$\mathrm{\omega_{d} = \sqrt{\omega_{n}^{2}-\zeta^{2}\omega_{n}^{2}}}$$At resonance, the amplitude of the system will be maximum when forced by a sinusoidal force of frequency equal to the resonant frequency.Resonant frequency:$$\mathrm{\omega_{d} = \sqrt{\omega_{n}^{2}-\zeta^{2}\omega_{n}^{2}}}$$$$\mathrm{\omega_{d} = \sqrt{(6.57)^{2}-(-2.88)^{2}} = 6.98 rad/s}$$Hence, the frequency of applied force at which resonance will occur is 6.98 rad/s.

The frequency of the applied force at which resonance will occur is ω = 2√5 rad/s.

To determine the frequency of the applied force at which resonance will occur, resonance happens when the frequency of the applied force matches the natural frequency of the system. The natural frequency can be determined using the formula:

ωn = √(K / M),

where ωn is the natural frequency, K is the spring constant, and M is the mass of the system.

Substituting the given values of K = 400 N/m and M = 20 kg into the equation, we can calculate the natural frequency ωn.

ωn = √(400 N/m / 20 kg) = √(20 rad/s²) = 2√5 rad/s.

Therefore, the frequency of the applied force at which resonance will occur is ω = 2√5 rad/s.

The correct question is given as,

M= 20kg

Fo = 90 N

ω = 6 rad/s

K = 400 N/m

C = 125 Ns/m

Determine the frequency of applied force at which resonance will occur?

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Newton's third law of motion states that for every action, there is an equal and opposite reaction. The action-reaction pair is used to describe the interaction between two objects. Therefore, when a car is speeding up as it drives along a level road, an action-reaction pair occurs.

An action-reaction pair is a pair of forces that are equal in strength and opposite in direction. When an object exerts a force on another object, the second object exerts an equal and opposite force on the first object. The action-reaction pair when a car speeds up as it drives along a level road can be explained as follows: Action force: The car exerts a force on the road in the forward direction. This is the action force. Reaction force: The road exerts a force on the car in the backward direction. This is the reaction force.

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Answer:

Bulb a

Explanation:

Explanation: It's really blurry and sideways

The initial velocity of the car just before the brakes were applied is 17.80 m/s

From the question given above, the following data were obtained:

The initial velocity of the car can be obtained as illustrated below:

v² = u² + 2as

0² = u² + (2 × -7.2 × 22)

0 = u² + (-316.8)

0 = u² - 316.8

Collect like terms

u² = 0 + 316.8

u² = 316.8

Take the square root of both side

u = √316.8

u = 17.80 m/s

Thus, we can conclude that the initial velocity of the car before the brakes were applied is 17.80 m/s

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The shift or change in a wave's frequency or wavelength brought on by the proximity of an emitting or reflecting sound source to the receiver. Therefore, the wavelength of the sound wave is 1.10 meters.

You may have noticed that the pitch of a fast-moving siren suddenly decreases as it passes past you. Because the pitch is higher, the siren initially appears to be approaching you. The siren's pitch drops as it moves away from you after passing you. The Doppler effect is being displayed here.

wavelength = speed / frequency

where wavelength is in units of meters, speed is in units of meters per second (m/s), and frequency is in units of Hertz (Hz).

For a sound wave with a speed of 680 m/s and a frequency of 220 Hz:

wavelength = speed / frequency

wavelength = 680 m/s / 220 Hz

wavelength = 3.09 m

Therefore, the wavelength of the sound wave is 3.09 meters.

For a sound wave with a speed of 960 m/s and a frequency of 875 Hz:

wavelength = speed / frequency

wavelength = 960 m/s / 875 Hz

wavelength = 1.10 m

Therefore, the wavelength of the sound wave is 1.10 meters.

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Answer:

I = 0.058 [A]

Explanation:

Ohm's law is defined as voltage equal to the product of current by resistance.

\(V =I*R\)

where:

V = voltage = 4.2 [V]

I = current [amp]

R = resistivity = 72 [ohm]

Now clearing I, we have:

\(I = V/R\\I = 4.2/72\\I = 0.058 [A]\)

Answer:

A. 0.058 A

Explanation:

Answer:

85 miles .

Explanation:

Displacement along the 110 South freeway = 260 - 150  =  110 miles

Displacement along the 110 North freeway = 150 - 175   = - 25 miles

Net displacement = 110 - 25 = 85 miles

So Joey's displacement from the 260 mile marker is 85 miles .

Answer:

Parallel Circuits

Explanation:

Loads can be operated on their own. If you had a series circuit, adding another light would dim the rest!

Answer:

35

Explanation:

total magnification = eyepiece lens x objective lens

TM = 10X x 25X

TM = 250X

a) Sketch the plate shape: we get a shape that resembles a trapezoid.

The plate shape is determined by the lines y = 2 - x² and y = 2x - 1. To sketch the plate shape, we can plot these two lines and shade the region in between them. The intersection points of the lines are found by solving the equations simultaneously:

2 - x² = 2x - 1

Simplifying, we get:

x² + 2x - 3 = 0

Factoring, we have:

(x - 1)(x + 3) = 0

So, x = 1 and x = -3. Plugging these values into the equations of the lines, we find the corresponding y-values:

For x = 1:

y = 2 - (1)² = 1

For x = -3:

y = 2(-3) - 1 = -7

Plotting these points and connecting them with the lines, we get a shape that resembles a trapezoid.

b) Total charge through the double integral:

To find the total charge Q, we need to integrate the charge density p(x, y) over the entire plate. We can express this as a double integral:

Q = ∬ p(x, y) dA

c) Reducing the double integral to repeated integrals: The limits of integration for x are the values of x that define the boundaries of the plate shape, which are -3 to 1.

Since the plate shape is described by the lines y = 2 - x² and y = 2x - 1, we can rewrite the double integral as a repeated integral by integrating with respect to x and y separately:

Q = ∫∫ p(x, y) dy dx

The limits of integration for y are from the lower curve y = 2 - x² to the upper curve y = 2x - 1. The limits of integration for x are the values of x that define the boundaries of the plate shape, which are -3 to 1.

d) Calculating the integral: The total charge Q of the thin plate is approximately 12.4 mC.

Now, we can evaluate the double integral to find the total charge Q:

Q = ∫(-3 to 1) ∫(2 - x² to 2x - 1) x²y dy dx

Performing the inner integral with respect to y first, we get:

Q = ∫(-3 to 1) [x²(y²/2 - y)] from 2 - x² to 2x - 1 dx

Simplifying the inner integral, we have:

Q = ∫(-3 to 1) [(x²/2)(2 - x²) - x²(2x - 1)] dx

Expanding and simplifying further, we get:

Q = ∫(-3 to 1) (x² - x⁴/2 - 4x³ + 2x²) dx

Integrating term by term, we have:

Q = [x³/3 - x⁵/10 - x⁴ + 2x³/3] from -3 to 1

Evaluating the integral at the limits, we get:

Q ≈ 12.4 mC (rounded to five significant figures)

Therefore, the total charge Q of the thin plate is approximately 12.4 mC.

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Answer:B

Explanation:

Being familiar with the different types of galaxies that exist in the universe is the action which might lead the scientists to develop new explanations about the universe. Thus, the correct option is D.

Many religious person, including many great scientists, hold that the God has created the universe and the various processes which are the driving physical and biological evolution and that these processes then resulted in the creation of the different galaxies, the solar system, and life on the planet Earth.

The Big Bang theory states that the universe came into being from a single, unimaginably hot and dense point more than about 13 billion years ago. This did not occur in an already existing space. Rather, it initiated from the expansion and cooling of space itself.

Therefore, the correct option is D.

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