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3 years ago

Could someone help me with this please?

Physics

2 answers:

ycow [4]3 years ago

6 0

Answer:

you could measure several properties of

the unknown liquid and compare them with the properties of known

substances. You might observe and measure such properties as color,

odor, texture, density, boiling point, and freezing point.

ratelena [41]3 years ago

6 0

Answer:

Compare the various physical properties of the substance.

Explanation:

One way is to compare the various physical properties of the substance. Let's take a look at those:

Identify the physical identity of each substance. That is, it the substance solid, or powder, or granulated (in small granules)

Next, the color of the substance will also help as a physical property. For example, is the substance bright, shiny, dull in color, etc.

Another property of the test can be the electrical conductance. For instance, does the substance conduct electricity or not.

Finally, test if the substance dissolves in a solvent, say water in this case. If it dissolves, it is likely to be a soluble salt. If not, then it is likely to be an insoluble salt.

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3 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

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3 years ago

Which of the following is a potential result of preparing appropriately before starting an experiment in a lab?

oksano4ka [1.4K]

Answer:

Your project goes well.

Explanation:

Because that's how it works.

6 0

3 years ago