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4 years ago

How and why has the Atomic Theory changed over time

1 answer:

3 0

In chemistry and physics, the atomic theory explains how our understanding of the atom has changed over time. Atoms were once thought to be the smallest pieces of matter. ... The first idea of the atom came from the Greek philosopher Democritus.

I hope this helps!

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Compare the foot of mollusks with the tube feet of echinoderms.

stira [4]

Mollusks live in fresh water, in marine environment, but also on land.

Echinoderms live only in water. This is the reason why they have developed different systems for moving. Mollusks have only singular muskullus foot for walking and Echinoderms have tube feet which they use for moving, as well for collecting and transporting food to their mouth.

4 0

4 years ago

Pls hurry worth 28 points <3

Kipish [7]

Answer:

Explanation:

I will GUESS that we're supposed to be looking at a plot of a position in time.

IF that is the case.

THEN the answer would be Point B because it has the steepest slope.

6 0

3 years ago

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

Now the intensity of the sound at the given location is related mathematically as:

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

As we know :

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

Now the velocity of mosquito for the same kinetic energy:

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

Using eq. (1)

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0

3 years ago

Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with therma

Sloan [31]

Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance ( $C_{HP}$ ) is given by

$C_{HP} = \dfrac{Q_{H}}{W_{in}}$

where, $Q_{H}$ is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature $T_{H}$ and $W_{in}$ is the required input and is given by $W_{in} = Q_{H} - Q_{L}$ , $Q_{L}$ being magnitude of heat transfer between cyclic device and low-temperature $T_{L}$ . Therefore, from above equation we can write,