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3 years ago

What happens to an object's gravitational potential energy as it falls from some height?

Physics

1 answer:

saveliy_v [14]3 years ago

3 0

Answer:

Decreases

Explanation:

As an object falls from some height, the gravitational potential energy of the object decreases.

Potential energy is the energy due to the position of a body.

The potential energy is mathematically expressed as:

Potential energy = mass x acceleration due to gravity x height

So, potential energy is directly proportional to mass and height.

The higher the mass and height, the more the potential energy.

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The second Law of Thermodynamics states that: A. spontaneous processes are characterized by the overall conversion of order to d

Georgia [21]

Answer:

Spontaneous processes are characterized by the overall conversion of order to disorder.

Explanation:

The second law of thermodynamics states that: A spontaneous process occurs only if there is an increase in entropy of a system and its surroundings.

Entropy, S, is a measure of the randomness or disorder of a system. It is measured in J/Kmol.

The change in entropy, ∆S = ∆H/T

Where ∆H = change in enthalpy, T = Temperature in Kelvin.

For,

I. An endothermic reaction, ∆S = positive (that is, ∆S is greater than zero), there is an increase in entropy, therefore, the reaction is spontaneous.

II. An exothermic reaction, ∆S = negative (that is, ∆S is less than zero) there is a decrease in entropy, so, the reaction is non-spontaneous.

III. A system at equilibrium, ∆S = 0.

Then,

The standard change in entropy of a reaction, ∆So reaction , is the difference in the standard entropies between products and reactants:

∆So reaction = n ∆Soproducts - m ∆Soreactants

Where, = sigma = sum of,

∆ = delta = change in,

n and m = stoichiometric coefficients of the products and reactants respectively.

Furthermore, the entropy of the system and surroundings is referred to as the entropy of the universe.

∆Suniverse = ∆Ssurroundings + ∆Ssystem.

Processes leading to an increase in entropy include melting, heating, vaporization, dissolving.

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3 years ago

The Nature of Sound?

bogdanovich [222]

Longitudinal wave hope this is right :))))))))))

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4 years ago

Read 2 more answers

Technician A says the parking brake cable adjustment should be performed before adjusting the rear brakes. Technician B says the

Firlakuza [10]

Answer:

Neither both are correct

Explanation:

The question involves the removal and installation procedure of cable for parking break. The answer for this could be found through the manual procedure for repairing parking break

Technician A is wrong because rear brakes should be adjusted before the parking cable adjustment is being made.

Technician B is wrong because the parking brake lever should be intact and secure at all clicks to allow maximum security. 15 clicks in this case should be the maximum number of clicks for the lever.

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3 years ago

The total distance from a house to a school to is 9.5 km. A student travels all the way from his house to the school and back to

marysya [2.9K]

Answer:

Explanation:

Displacement is a vector from initial to final point. Because initial and final point are the same, so displacement is 0.

6 0

3 years ago

1) Consider a source particle of charge qS=9 C located at (−9,5) [distances in meters]. Find the electric field vector at the ta

xeze [42]

Answer:

1) E = 2.25 i^+ 0.809j^) 10⁹ N / C , 2) E = 2.39 10⁹ N / C , 3) θ = 19.8º , 4) F = 19.12 10⁹ N , 5) E = (1.32 i^+ 3.56 j^) 109 N/C

Explanation:

1) The equation for the electric field is

E = k q / r²

Where K is the Coulomb constant that is worth 8.99 10⁹ N m² /C², q is the load and r is the distance of the load to the test point

Since the electric field is a vector magnitude, we can find its component

X axis

Ex = k q / x²

where the distance on the axis is

x = √ (X₂-x₁)²

x = √ (-15 + 9)² = 6 m

Eₓ = 8.99 10⁹ 9/6²

Eₓ = 2.25 10⁹ N /C

Y Axis

y = √ (y₂-y₁)² = √ (15-5)² = 10 m

$E_{y}$ = 8.99 10⁹ 9/10²

$E_{y}$ = 0.809 10⁹ N / C

E = Eₓ i^+ $E_{y}$ j^

E = 2.25 i^+ 0.809j^) 10⁹ N / C

2) the magnitude can be found using the Pythagorean triangle

E = √ (Eₓ² + $E_{y}$ ²)

E = √ (2.25² + 0.809²) 10⁹

E = 2.39 10⁹ N / C

3) to find the angle let's use trigonometry

tan θ = $E_{y}$ / Eₓ

θ = tan⁻¹ $E_{y}$ / Eₓ

θ = tan⁻¹ (0.809 / 2.25)

θ = 19.8º

Regarding the positive side of the x axis

4) a charge q2 = 8C is placed, let's calculate the force

F = q E

F = 8 2.39 10⁹

F = 19.12 10⁹ N

5) The total electric field at the origin, let's look for its components

q₁ = 9C

r₁ = -9 i ^ + 5 j ^

q₂ = 8 C

r₂ = -15 i ^ + 15 j ^

X axis

Eₓ = E₁ₓ + E₂ₓ

Eₓ = k q₁ / Dx₁² + k q₂ / Dx₂²

Eₓ = 8.99 10⁹ (9 / (9-0)² + 8 / (15-0)²)

Eₓ = 1.32 109 N / A

Y Axis

$E_{y}$ = $E_{1y}$ + $E_{2y}$

$E_{y}$ = k q₁ / Δy₁² + k q₂ / Δy₂²

$E_{y}$ = 8.99 109 (9/5² + 8/15²)

$E_{y}$ = 3.56 109 N / A

E = (1.32 i^+ 3.56 j^) 109 N/C

7 0

4 years ago

What happens to an object's gravitational potential energy as it falls from some height?​

What happens to an object's gravitational potential energy as it falls from some height?