15) What is the momentum of a 280 kg motorcycle moving with a velocity of 15 m/s?

Generally, what will happen to most liquids if the temperature is increased?

GalinKa [24]

Temperature doesn't really affect solubility on liquids so it can only be D besides it's already a liquid....

5 0

3 years ago

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Which feature or process is unique to nuclear power plants when compared to conventional coal-burning power plants?

Mazyrski [523]

Answer:

Control rods

Explanation:

Control rods is unique to nuclear power plants as they are mainly used in nuclear reactors to stabilize the rate of fission of plutonium or uranium. They are usually composed of chemical elements such as cadmium, silver, boron, and indium.

5 0

2 years ago

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The amount of matter in an object is its _____.

Andrew [12]

The amount of matter in an object ismass....anything that occupies spaca and has weight is called matter.....

7 0

2 years ago

Write any two different between work and power?

attashe74 [19]

Answer:

1. Work Is a type of Physical Activity

2. Power is Basically Having control of Society

Explanation:

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2 years ago

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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