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tigry1 [53]
4 years ago
15

One advantage of light microscopy over transmission electron microscopy is that One advantage of light microscopy over transmiss

ion electron microscopy is that light microscopy provides for higher magnification than transmission electron microscopy. specimen preparation for light microcopy does not produce artifacts. light microscopy provides for higher resolving power than transmission electron microscopy. light microscopy allows one to view dynamic processes in living cells.
Physics
1 answer:
il63 [147K]4 years ago
4 0

Answer:

d. light microscopy allows one to view dynamic processes in living cells.

Explanation:

One advantage of light microscopy over transmission electron microscopy is that light microscopy allows one to view dynamic processes in living cells.

Electron microscope differ from the light microscope because electron micoscope produce the image of the specimen using electron beam and not light.  Electron microscopy samples must be kept in vacuum, this means live cells can not be imaged.

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The hydraulic oil in a car lift has a density of 8.53 x 102 kg/m3. The weight of the input piston is negligible. The radii of th
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Answer:

(a) the input force is 36.56 N

(b) the input force is 37.49 N

Explanation:

Given;

density of hydraulic oil, ρ =  8.53 x 10² kg/m³

radius of plunger, r₁ = 0.135 m

radius of piston, r₂ = 5.43 x 10⁻³ m

Part (a) The input force needed to support 22600-N weight, when the bottom surfaces of the piston and plunger are at the same level;

P =\frac{F}{A}

Where;

P is pressure

F is force

A is circular area = πr²

\frac{F_1}{A_1} =\frac{F_2}{A_2} \\\\F_2 = \frac{F_1*A_2}{A_1} =\frac{F_1* \pi r_2^2}{\pi r_1^2} = \frac{F_1*  r_2^2}{ r_1^2} \\\\F_2 = \frac{22600*(5.43*10^{-3})^2 }{(0.135)^2}\\\\F_2 = 36.56 \ N

Part (b) The input force needed to support 22600-N weight, when the  bottom surface of the output plunger is 1.20 m above that of the input plunger

P_2 = P_1 + \rho gh

But, F = PA  and  A = πr²

F_2 = F_1(\frac{A_2}{A_1} ) + \rho gh*A_2\\\\F_2 = F_1(\frac{r_2^2}{r_1^2} )+\rho gh(\pi r_2^2)\\\\F_2 = 22600(\frac{5.43*10^{-3}}{0.135})^2 \ + 853*9.8*1.2*\pi (5.43*10^{-3})^2\\\\F_2=36.56 + 0.93\\\\F_2 = 37.49 \ N

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cochlea is part of the inner ear. have a great day my friend

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