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3 years ago

What could you do to change the volume of a gas?

1 answer:

8 0

The ONLY way to change the volume of a sample of gas is to transfer it to a container with different volume.
Simply changing its temperature or pressure in the same jar won't do it. Any amount of gas always fills whatever container you keep it in.

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A pulley is able to lift a mass of 25 kg 0.30 m with an applied force of 50 N over a distance of 1.5 m. What is the ideal mechan

spin [16.1K]

The ideal mechanical advantage of the pulley system is 3

7 0

3 years ago

Question 11 (1 point)

Ainat [17]

Answer:

gravitational potential energy.

Explanation:

Gravitational potential energy (GPE) can be defined as an energy possessed by an object or body due to its position above the earth surface.

Mathematically, gravitational potential energy is given by the formula;

$G.P.E = mgh$

Where,

G.P.E represents gravitational potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

This ultimately implies that, anytime there is height, the object must have gravitational potential energy.

Hence, an object possesses gravitational potential energy due to its height (position) and the earth's gravitational force.

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2 years ago

What is imperceptible human motion?

castortr0y [4]

Unable to be noticed or not felt because of feeling slight

3 0

3 years ago

The data table shows some data related to the Sun and the planets in our solar system.

AlexFokin [52]

You have selected the correct answer and blobbed over it with your pencil.

I assume you must have looked at Saturn's average distance, found 1427,
divided that number by 6, got 237 and change, then looked at the others,
and found that 228 was the only one that's anywhere close.

8 0

3 years ago

Read 2 more answers

A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t

sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, $\rho=2.44\times 10^{-8}\ \Omega-m$

Resistance in terms of resistivity is given by :

$R=\dfrac{\rho l}{A}$

Also, V = IR

So,

$\dfrac{V}{I}=\dfrac{\rho l}{A}$

A is area of wire,

$\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}$ , r is radius, r = d/2 (diameter=d)

$\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m$

Out of four option, near option is (C) 17 μm.

6 0

3 years ago