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3 years ago

I’m Not Sure What To Do ..

Physics

1 answer:

den301095 [7]3 years ago

4 0

Answer:solve it

Explanation:

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3 years ago

PLEASE HELP!

luda_lava [24]

Answer:

When a body is placed on a table top, it exerts a force equal to its weight downwards on the table top but does not move or fall. (i) Name the force exerted ...

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3 years ago

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.

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

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3 years ago

If the resulting trajectory of the charged particle is a circle,what is , the angular frequency of the circular motion?Express i

Maru [420]

Answer:

ω = B₀q/m

Explanation:

The centripetal force on the charge equal the magnetic force on the charge. (Since the trajectory is a circle). So, mrω² = B₀qv.

v = rω The speed of the charge and r = radius of path

mrω² = B₀qrω

ω = B₀q/m

The angular frequency ω = B₀q/m

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3 years ago