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1 year ago

Using a _____ microscope, cells appear as translucent objects against a bright background

Physics

1 answer:

zloy xaker [14]1 year ago

5 0

Cells can be observed under a brightfield microscope as translucent objects against a vivid background.

<h3>By translucent items, what do you mean?</h3>

Objects that are translucent allow some light to pass through them. Translucent materials include certain polymers and frosted glass. Only a portion of the light that strikes transparent materials goes through. It takes the materials some time for the light to flow through them. Waxed paper, thin plastic sheets, butter paper, and vegetable oil are a few examples of translucent materials. This mnemonic should help you remember when to use transparent because transparent objects are clear and light may readily pass through them. The thing would be translucent if light still enters it, but it is less noticeable. An opaque object is one through which no light can flow.

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A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

$v=\sqrt{\frac{TL}{M}}$ (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

$v= \lambda f$ (2)

We can equate expression (1) and (2):

$\sqrt{\frac{TL}{M}}$ = $\lambda f$

Solving for T

$T= \frac{M(\lambda f)^2}{L}$ (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic $N_{harmonic}$ is:

$\lambda_{harmonic}=\frac{2l}{N_{harmonic}}$

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

$\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m$

We can now find T on (3) using all the values we have:

$T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}$

$T=75.22 N$

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