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2 years ago

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

Physics

1 answer:

saveliy_v [14]2 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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A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper pla

Hatshy [7]

Answer:

The time rate of change of the electric field between the plates is $\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}$

Explanation:

From the question we are told that

The radius is $r = 2.8 \ cm = 0.028 \ m$

The distance of separation is $d = 1.1 \ mm = 0.0011 \ m$

The current is $I = 5 \ A$

Generally the electric field generated is mathematically represented as

$E = \frac{q }{ \pi * r^2 \epsilon_o }$

Where $\epsilon_o$ is the permitivity of free space with a value

$\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

So the time rate of change of the electric field between the plates is mathematically represented as

$\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }$

But $\frac{q}{t } = I$

$\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }$

substituting values

$\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }$

$\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}$

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3 years ago

A heat engine operating at steady state delivers a power output of 15.5 hp. The engine receives energy by heat transfer in the a

Anna007 [38]

Answer:

w_cycle = 13.68 KJ / cycle

Explanation:

Given:

- Steady power out-put W_out = 15.5 hp

- System executes 50 cycles/min

Find:

Determine the amount of work that is delivered during each cycle in kilo-joules

Solution:

- First we will convert the steady output into KW as follows:

W_out = 15.5 hp

The conversion factor is 1.36 KW per hp.

W_out = 15.5 [hp] * [KW] / 1.36 [hp]

W_out = 11.4 [KW] or [KJ/s]

- Given that there are 50 cycles in 60 sec. We will use direct proportionality to calculate the number of cycles per second:

50 cycles -----------> 60 s

x cycles -----------> 1 s

x = 50/60 = 0.8333 cycles /s

- Next we will compute the amount of work done per cycle:

w_cycle = W_out / x

w_cycle = 11.4 [KJ/s] * [s] / 0.8333 [cycles]

w_cycle = 13.68 KJ / cycle

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2 years ago

Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c

shtirl [24]

Responder: A.) 24.5m / s B.) 30.625m C.) 89.375m

Explicación:

Dado lo siguiente:

Altura desde la cual se cae el cuerpo = 120 m

Tiempo (t) = 2.5s

A.) La velocidad que toma:

El cuerpo cayó desde una altura;

velocidad inicial (u) = 0

Para calcular v:

V = u + en

Donde a = aceleración debido a la gravedad = 9.8m / s

v = 0 + (9.8) (2.5)

v = 24.5 m / s

B) Cuánto ha disminuido.

Usando la ecuación de movimiento:

S = ut + 0.5at ^ 2

Donde S = distancia

S = 0 × 2.5 + 0.5 (9.8) (2.5 ^ 2)

S = 0 + 0.5 (9.8) (6.25)

S = 30.625 m

Esta es la distancia recorrida después de 2.5 segundos Altura o distancia ha disminuido en 30.625 m

C.) ¿CUÁNTO FALTA? Por lo tanto, 120m - 30.625m = 89.375m

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3 years ago

2. Two spherical objects with radii 1 m have masses of 1.5 kg and 8.5 kg. They are separated by a

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Answer:

They are separated by a distance of 18 m. Find the gravitational attraction between them. r= 1 m r= 1 m 18 m. Mass = 1.5 kg. Mass = 8.5 kg.

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2 years ago

Which of these statements are true?

Juliette [100K]

D.Neither High- nor low-pressure systems lead to rain

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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?