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Rzqust [24]
3 years ago
9

How do Leeuwenhoek’s observations of animalcules compare to Hooke’s observations of cells in the cork?

Physics
1 answer:
fomenos3 years ago
8 0

Answer:

Robert Hooke

Was the first to use the word "cell"

Observed cork cells

Anton van Leeuwenhoek

Observed "animalcules"

Used polished lens .

Explanation:

Anton van Leeuwenhoek is known as father of microbiology. He is credited to improve the quality of lens in microscope. His first observation of organisms called animalcules.

He is credited to have build microscope that could get magnified by 200 times. He used word animalcules for small organisms from pond water when first observed in microscope. He discovered protozoa and named it animalcules".

Robert Hooke is famed for discovering cell from a cork of plant. He observed a compartment or honey comb like divisions when observed these cork cells under the microscope and named it cell. He was only able to see the cell wall as the cork cells are dead cells.

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Answer:

330.4 N and  95.2^{\circ}counterclocwise to the x direction

Explanation:

Sum of forces in vertical are equal, let movement to right and upwards be positive while left and downwards be negative

Net force in horizontal direction is 85-115=-30 N

Net force in vertical direction is 565-236=329 N

Resultant force=\sqrt {(-30)^{2}+329^{2}}=330.3649497 N\approx 330.4 N

Direction=tan^{-1}\frac {329}{-30}=-84.78986932\approx -84.8^{\circ}

180-84.8=95.2

Therefore, direction is 95.2^{\circ}counterclocwise to the x direction

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4 years ago
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3 years ago
Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area
yaroslaw [1]

Answer:

5.63\cdot 10^{-6} C

Explanation:

The capacitor of a parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

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d is the separation between the plates

\epsilon_0 is the vacuum permittivity

The energy stored in a capacitor instead is given by

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Substituting the expression we found for C inside the last formula,

U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}

And re-arranging it

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Now if we substitute

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We find the charge stored on the capacitor:

Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C

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