What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure

Physics

1 answer:

mezya [45]3 years ago

8 0

Answer:

this is a no brainer

Explanation:

As air pressure in an area increases, the density of the gas particles in that area increases.

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Calculate the current in a circuit if 500 C of charge passes through it in 10 minutes. *

Sloan [31]

Answer:

Choice a. Approximately $0.83\; \rm A$ on average.

Explanation:

The electric current through a wire is the rate at which electric charge flows through a cross-section of this wire.

Assume that electric charge of size $Q$ flowed through a wire cross-section over a period of time $t$ . The average current in that wire would be:

$\displaystyle I = \frac{Q}{t}$ .

For this question:

$Q = 500\; \rm C$ , whereas
$t = 10\; \rm \text{minutes}$ .

Therefore, the average current in this circuit would be:

$\displaystyle I = \frac{Q}{t} = \frac{500\; \rm C}{10\; \text{minutes}} = 50\; \rm C /\text{minute}$ .

However, the units in the choices are all in $\rm A$ (for Amperes.) One Ampere is equal to one $\rm C / \text{second}$ . It will take some unit conversations to change the unit of $I = 50\; \rm C/ \text{minute}$ (coulombs-per-minute) to coulombs-per-second.

$\begin{aligned}I &= 50\; \rm C/ \text{minute} \\ &= \frac{50\; \rm C}{1\; \rm \text{minute}} \times \frac{1 \; \text{minute}}{60\; \rm \text{seconds}} \approx 0.83\; \rm C/ \text{second} = 0.83 \; \rm A\end{aligned}$ .