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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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PE is mgh in this context.
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Which best describes electrical energy? a. energy released by a chemical reaction b. energy produced by flow of electric charge
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4 0
3 years ago
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The near point of an eye is 56.0 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan
irakobra [83]

Answer:

Explanation:

Near point = 56 cm .

near point of healthy person = 25 cm

person suffers from long sightedness

convex lens will be required .

object distance u  = 25 cm

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both will be negative as both are in front of the lens.

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I/v - 1 / u = 1/f

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4 0
4 years ago
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Vika [28.1K]

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5 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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