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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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Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago

Two identical objects A and B fall from rest from different heights to the ground. If object B takes twice as long as A to reach

aivan3 [116]

I believe this ratio is 4:1 due to the inverse square law

4 0

2 years ago

Read 2 more answers

A 5kg bucket hangs from a ceiling on a rope. A student attaches a spring scale to the buckets handle and pulls horizontally on t

7nadin3 [17]

I don’t know what the angle is in your diagram so I used the angle from the vertical.

6 0

2 years ago

A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float

storchak [24]

Answer:

t ’= $\frac{1450}{0.6499 + 2 v_r}$ , v_r = 1 m/s t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

v_b - v_r = d / t

By the time the boat goes down the river

v_b + v_r = d '/ t'

let's subtract the equations

2 v_r = d ’/ t’ - d / t

d ’/ t’ = 2v_r + d / t

$t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }$

In the exercise they tell us

d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

t ’= $\frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \ + 2 \ v_r}$

t ’= $\frac{1450}{0.6499 + 2 v_r}$

In order to complete the calculation, we must assume a river speed

v_r = 1 m / s

let's calculate

t ’= $\frac{ 1450}{ 0.6499 + 2 \ 1}$

t ’= 547.19 s

8 0

2 years ago

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels.

shutvik [7]

Answer:

a = 6.53 m/s^2

v = 11.5689 m/s

Explanation:

Given data:

engine power is 217 hp

70 % power reached to wheel

total mass ( car + driver) is 1530 kg

from the data given

2/3 rd of weight is over the wheel

w = 2/3rd mg

maximum force

$F = \mu W$

we know that F = ma

$ma = \mu (2/3 mg)$

$a_{max} = 2/3(1.00) (9.8) = 6.53 m/s^2$

the new power is $p = 70\% P_[max} = 0.7 P_{max}$

$P =f_{max} v$

$0.7P_{max} = ma_{max} v$

solving for speed v

$v =0.7 \times \frac{P_{max}}{ma_{max}}$

$v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}$

v = 11.5689 m/s

7 0

2 years ago