As a battery is used to charge a capacitor, does the overall charge inside the battery get smaller, greater, or stay the same?

A gas occupies a volume of 1.0 m3 in a cylinder at a pressure of 120kPa. A piston compresses the gas until the volume is 0.25m3,

Hoochie [10]

Answer:

Approximately $480\; \rm kPa$ , assuming that this gas is an ideal gas.

Explanation:

Let $V(\text{Initial})$ and $P(\text{Initial})$ denote the volume and pressure of this gas before the compression.
Let $V(\text{Final})$ and $P(\text{Final})$ denote the volume and pressure of this gas after the compression.

By Boyle's Law, the pressure of a sealed ideal gas at constant temperature will be inversely proportional to its volume. Assume that this gas is ideal. By this ideal gas law:

$\displaystyle \frac{P(\text{Final})}{P(\text{Initial})} = \frac{V(\text{Initial})}{V(\text{Final})}$ .

Note that in Boyle's Law, $P$ is inversely proportional to $V$ . Therefore, on the two sides of this equation, "final" and "initial" are on different sides of the fraction bar.

For this particular question:

$V(\text{initial}) = 1.0\; \rm m^3$ .
$P(\text{Initial}) = 120\; \rm kPa$ .

$V(\text{final}) = 0.25\; \rm m^3$ .
The pressure after compression, $P(\text{Final})$ , needs to be found.

Rearrange the equation to obtain:

$\displaystyle P(\text{Final}) = \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial})$ .

Before doing any calculation, think whether the pressure of this gas will go up or down. Since the gas is compressed, collisions between its particles and the container will become more frequent. Hence, the pressure of this gas should increase.

$\begin{aligned}P(\text{Final}) &= \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial})\\ &= \frac{1.0\; \rm m^{3}}{0.25\; \rm m^{3}} \times 120\; \rm kPa = 480\; \rm kPa\end{aligned}$ .

4 0

3 years ago

Energy balance in the sun refers to a balance between _________.

lubasha [3.4K]

The amount of energy the Sun radiates into space and the amount of energy that reaches Earth.

6 0

3 years ago

A car, starting from rest, accelerates in a straight-line path at a constant rate of 2.0 m/s2. How far will the car travel in 12

Kryger [21]

Same formula as the last question. x = vt + (1/2)at^2. In this case, v = 0, t = 12, and a = 2.0. Plug in the values and solve for x (which is change in position) x = (0)(12) + (1/2)(2.0)(12^2) x = (1/2)(2.0)(144) x = (1)(144) x = 144 So the car will travel 144 meters in 12 seconds.

8 0

2 years ago

Given velocity, what is the formula for acceleration?

Bogdan [553]

Answer:

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

3 0

2 years ago

A heavy object falls with the acceleration as a light object during free fall. Why?

Lyrx [107]

A heavy object falls with the acceleration as a light object during free fall because of acceleration due to gravity.

Explanation:

A motion can be termed as free fall when the object is completely under the influence of gravity. So in this case, no other force will be acting on the object other than the gravitational force. As the gravity influences the object in free fall, the acceleration attained by any object in free fall is same. And this acceleration is termed as acceleration due to gravity. Since, the gravity gives the acceleration to every object experiencing free fall, then the acceleration should be constant. Thus, a heavy object and a light object will have same acceleration during free fall.

6 0

2 years ago