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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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Apply Newton's second law to the person's motion:

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Given values:

m = 50.8kg, a = 3.50m/s²

Plug in and solve for F:

F = 50.8(3.50)

F = 178N

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What movements of water has the greatest effect on the growth of producers
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To drive a car at a constant velocity, you
kipiarov [429]

Answer:

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2 years ago
How is a quarter able to role on its edge?
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5 0
4 years ago
A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
MissTica

Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )

\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )

\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )

To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

5 0
4 years ago
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