Controlled variables

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

7 0

Answer:

A control variable in scientific experimentation is an experimental element which is constant and unchanged throughout the course of the investigation.

Explanation:

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7. Copper can be coated on the surface of iron but not on silver? why

Lisa [10]

Answer:

Silver has greater electronegativity (pull on electrons) than copper, so it will be reduced rather than oxidized. Silver ions will plate out on copper metal. Sir the coating of iron with copper surface can be referred as IRON-COPPER PLATING.

Explanation:

Iron is used for the electroplating of so copper because iron falls above copper in the electrochemical series whereas silver will fall off below the copper in electrochemical series that makes it not reliable for the electroplating purpose

4 0

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.

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