A closed system consists of a pendulum that is swinging back and forth. If
the pendulum's kinetic energy decreases, what else must happen to the
energy of the system?
a
A. Its gravitational potential energy must decrease.
B. Its total mechanical energy must increase.
O C. Its total mechanical energy must decrease.
O D. Its gravitational potential energy must increase.

Answers

Answer 1

Answer:

The correct choice is D. Its gravitational potential energy must increase

Explanation:

Conservation of Mechanical Energy

The total amount of mechanical energy, in a closed system in the absence of dissipative forces like friction or air resistance, remains constant.

This means that energy cannot disappear or appear and that potential energy can become kinetic energy or vice versa.

In a closed system like a pendulum, two types of energies are considered: Gravitational potential (U) and kinetic (K). Thus, the sum of both energies must remain constant in time.

Suppose the pendulum is at a state where U=150 J, and K=350 J. The total mechanical energy is:

M = 150 J + 350 J = 500 J

If the kinetic energy decreases to a new value, say K = 200 J, then the gravitational potential must increase to compensate for this new condition, that is: U = 300 J

The correct choice is D. Its gravitational potential energy must increase

Answer 2

Taking into account the definition of kinetic, potencial and mechanical energy, the correct answer is option D:  If  the pendulum's kinetic energy decreases Its gravitational potential energy must increase.

Kinetic energy

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and at rest, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its state of rest by applying a force.

Potential energy

On the other hand, potential energy is the energy that measures the ability of a system to perform work based on its position. In other words, this is the energy that a body has at a certain height above the ground.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the force of gravity.

Mechanical energy

Finally, mechanical energy is that which a body or a system obtains as a result of the speed of its movement or its specific position, and which is capable of producing mechanical work. Then:

Potential energy + kinetic energy = total mechanical energy

The principle of conservation of mechanical energy indicates that the mechanical energy of a body remains constant when all the forces acting on it are conservative (a force is conservative when the work it does on a body depends only on the initial and final points and not the path taken to get from one to the other.)

Therefore, if the potential energy decreases, the kinetic energy will increase. In the same way, if the kinetics decreases, the potential energy will increase.

This case

This principle can be applied in this case. The mechanical energy of the pendulum remains constant when all the forces acting on it are conservative.

So, the correct answer is option D:  If  the pendulum's kinetic energy decreases Its gravitational potential energy must increase.

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Answer:

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i know this im in 6th grade

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Answer:

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Answer:

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Answers

Answer:

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Explanation:

From Physics we get that resultant velocity of an airplane is the sum of an absolute velocity and a relative velocity, that is:

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Where:

[tex]\vec v_{W}[/tex] - Wind velocity, measured in meters per second.

[tex]\vec v_{A/W}[/tex] - Airplance velocity relative to wind, measured in meters per second.

[tex]\vec v_{A}[/tex] - Airplane velocity, measured in meters per second.

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Answer:

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Answer:

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Source: Trust me bro

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Donations to non-profit organizations are considered charitable contributions. Donations may take the form of cash, non-monetary goods, in-kind services, or volunteer time. According to IRS rules, the donation may be tax-deductible.

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Answer:

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Answer:

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Answer:

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Answers

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Answers

Answer: Option B.

Explanation:

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Answer:

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answer:

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explanation:

Answer: Heat

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On an icy day, you worry about parking your car in your driveway, which has an incline of 12º. Your neighbor's driveway has an incline of 9.0º, and the driveway across the street is at 6.0º. The coefficient of static friction between tire rubber and ice is 0.15. Which driveway(s) will be safe to park in?

Answers

Answer:

Driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.

Explanation:

Let suppose that car is represented by a particle, then we proceed to construct its free body diagram and corresponding equations of equilibrium:

[tex]\Sigma F_{x'} = f - W\cdot \sin \theta = 0[/tex] (Eq. 1)

[tex]\Sigma F_{y'} = N-W\cdot \cos \theta = 0[/tex] (Eq. 2)

Where:

[tex]f[/tex] - Static friction force, measured in newtons.

[tex]N[/tex] - Normal force on the car from the ground, measured in newtons.

[tex]W[/tex] - Weight of the car, measured in newtons.

[tex]\theta[/tex] - Driveway inclination, measured in sexagesimal degrees.

By applying definitions of maximum static friction force and weight, we expand the system of equations presented above:

[tex]\mu_{s}\cdot N -m\cdot g\cdot \sin \theta = 0[/tex] (Eq. 1b)

[tex]N - m\cdot g \cdot \cos \theta = 0[/tex] (Eq. 2b)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

[tex]m[/tex] - Mass of the car, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

Then, we substitute normal force and simplify the resulting expression:

[tex]\mu_{s}\cdot m\cdot g \cdot \cos \theta -m\cdot g \cdot \sin \theta = 0[/tex]

[tex]\mu_{s} = \tan \theta[/tex]

[tex]\theta = \tan^{-1}\mu_{s}[/tex] (Eq. 3)

Based on such result, we can conclude that if driveway inclination is greater than value reported, then car shall not be safe in case of parking. Our reference angle is: ([tex]\mu_{s} = 0.15[/tex])

[tex]\theta = \tan^{-1} 0.15[/tex]

[tex]\theta \approx 8.531^{\circ}[/tex]

By direct comparison, we find that driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.

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Answers

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