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grandymaker [24]
3 years ago
8

Which of the phases below are able to be compressed? A) Liquid only. B) Liquid and Gas. C) Solid only. D) Gas only

Physics
2 answers:
noname [10]3 years ago
6 0
Answer - B
Liquid and Gas can be compressed because of the freely moving particles in them. Solids cannot be compressed as their structure is fixed.
amid [387]3 years ago
3 0
The phrase below that is able to be compressed is D. gas only.
Although it is possible to compress liquids, it requires a great deal of pressure to accomplish a little compression, which is why liquids and solids are often referred to as incompressible. 
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Answer:

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Compare the geology of Callisto, Ganymede, and Titan.
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Which of the following statements is TRUE about updating the exposure control plan?
iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

<em>Updates must have the  reflection of changes in tasks as well in procedures.</em>

<em>Updates must reflect changes in positions that affect occupational exposure.</em>

<em>Updates must have the cost of PPE that is needed and  necessary to reduce exposure</em>

An exposure control plan can be regarded as  the framework for compliance between the employer and the workers.

  • This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.

  • This plan gives hope to workers in term of protection when working with their Employer.

  • There are some elements that is associated with  Exposure Control Plan, and theses are;
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2 years ago
What needed to be present in marine mud to form fossils fuels?
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3 years ago
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A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
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