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3 years ago

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19. Assume you have been chewing a piece of bread for 5 or 6 minutes. How would you expect its taste to change during this inter

val? Why?

1 answer:

I am Lyosha [343]3 years ago

7 0

Well 1 minute in it would taste like normal bread but 2-3 would the bread will break down and then get really soggy and 4-6 it would be into really tiny peices or almost gone and really wet from your saliva

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Which particle would have the slowest rate of deposition? A.round particle B.very large particle C.particle with sharp ends D.pa

lisabon 2012 [21]

The particle with sharp ends have the slowest rate of deposition

Answer: Option C

<u>Explanation:</u>

As per aerosol physics, deposition is a process where aerosol particles accumulate or settle on solid surfaces. Thereby, it reduces the concentration of particles in the air. Deposition velocity (rate of deposition) defines from F = vc, where v is deposition rate, F denotes flux density and c refers concentration.

Deposition velocity is slowest for particles of intermediate-sized particles because the frictional force offers resistance to the flow. Density is directly proportional to the deposition rate so clearly shows that high-density particles settle faster. Due to friction, round and large-sized particles deposit faster than oval/flattened sediments.

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3 years ago

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How are the Earth's crust and jigsaw alike?

Neporo4naja [7]

The earth's surface is composed tectonic plates of basaltic oceanic crust with lighter granitic continental crust floating in isostatic balance on the oceanic crust.<span>
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3 years ago

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A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig

SCORPION-xisa [38]

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : $f=20\ cm$

Height of object : $h=1\ cm$

Object distabce from lens : $u=-10\ cm$

Using lens formula: $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ , we get

$\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}$ , where v = image distance from the lens.

On solving aboive equation , we get

$\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm$

Formula of Magnification : $m=\dfrac{v}{u}=\dfrac{h'}{h}$ , where h' is the height of image.

Put value of u, v and h in it , we get

$\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm$

Hence, the height of the image is 2 cm.

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3 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

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3 years ago

3.5-m-diameter merry-go-round is rotating freely with an angular velocity of 0.78 rad/s . Its total moment of inertia is 1710 kg

Monica [59]

I dont get the question your asking

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3 years ago