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2 years ago

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

Physics

1 answer:

fomenos2 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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a bus is moving at 22m/s [E] for 12s. Then the bus driver slows down at 1.2m/s2 [W] until the bus stops. Determine the total dis

KatRina [158]

The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:

d = vt = (22 m/s)(12 s) = 264 m

For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²

Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m

Thus,
Total Displacement = 264 m + 201.67 m = 465.67 or approximately 4.7×10² m.

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1 year ago

A solenoid that is 35 cm long and contains 450 circular coils 2.0 cm in diameter carries a 1.75-A current. (a) What is the magne

Taya2010 [7]

Answer:

Explanation:

a ) No of turns per metre

n = 450 / .35

= 1285.71

Magnetic field inside the solenoid

B = μ₀ n I

Where I is current

B = 4π x 10⁻⁷ x 1285.71 x 1.75

= 28.26 x 10⁻⁴ T

This is the uniform magnetic field inside the solenoid.

b )

Magnetic field around a very long wire at a distance d is given by the expression

B = ( μ₀ /4π ) X 2I / d

= 10⁻⁷ x 2 x ( 1.75 / .01 )

= .35 x 10⁻⁴ T

In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .

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2 years ago

Read 2 more answers

Explain how cognitive psychologists combine traditional conditioning models with cognitive processes.

user100 [1]

Behaviorists generally claimed that conditioning occurred without thinking or reasoning ans was simply a result of consequences or reinforcement. Cognitive psychologists demonstrated that thinking and reasoning (cognition) influences the conditioning processes and that many behaviors that are conditioned depend on the type of cognitive reasoning that occurs during conditioning. Therefore, as one is being conditioned to respond to environmental stimuli or is responding to a consequence, they are also pondering and thinking about the process occuring. Cognition is often the reason individuals are not all conditioned in the same manner.

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1 year ago

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The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This sp

Delvig [45]

Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g and $60^\circ$ for $\theta$ in above equation.

$t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\$

Spitted water will travel $0.80{\rm{ m}}$ horizontally.

Displacement of water in this time period is

$x = vt\cos \theta$

Substitute $\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}$ for $t\rm 60^\circ[tex] for [tex]\theta$ and $0.80{\rm{ m}}$ for x in above equation.

$\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

${v_v} = v\sin \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

Horizontal component of the velocity is,

${v_h} = v\cos \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

When horizontal $({0.60\;{\rm{m}}}$ distance away from the fish.

The time of flight for distance (d) is ,

$t = \frac{d}{{{v_h}}}$

Substitute $0.60\;{\rm{m}}$ for d and $1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_h}$ in equation $t = \frac{d}{{{v_h}}}$

$\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\$

Distance of the insect above the surface is,

$s = {v_v}t + \frac{1}{2}g{t^2}$

Substitute $2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_v}$ and $0.3987{\rm{ s}}$ for t and $- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g in above equation.

$\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\$

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If you subscribe I’ll answer QF Aotrx

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