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sesenic [268]
3 years ago
12

Students were asked to place a mint in their mouths and determine how long it took for the mint to dissolve. Students were asked

to leave a whole mint in their mouth and let it dissolve. Other student groups varied how the mint was treated. One group was asked to swirl the whole mint around in their mouth; another group cut the mint in half, and last group broke the mint into small pieces. The results are displayed in the table.
What conclusion can be drawn from the results?
A) Surface area did not play a role in this activity.
B) Agitation of the mint increased the dissolving time.
C) As surface area increased, the dissolving time decreased.
D) The heat of the mouth helped to decrease the time for the mint to dissolve.

Solute Time to dissolve (seconds)
whole mint 192
mint cut in half 108
swirled mint 76
mint cut into small pieces 38
Physics
2 answers:
Arte-miy333 [17]3 years ago
8 0
The correct answer is C, the dissolving time will decrease due to the increased surface area.

lesya692 [45]3 years ago
4 0

Answer: C) As surface area increased, the dissolving time decreased.  

Surface area is the area which is exposed for a activity or process to take place. In this case surface of area being judged upon on two cases when students were asked to keep the whole mint in their mouth as such and in other when they were asked to swirl it in the mouth. Swirling inside the mouth will increase the surface area of the mint for dissolution inside the mouth. Therefore, as surface area increased, the dissolution time decreased.

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Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, T_{b} = 75^{\circ}C

Air temperature, T_{infty} = 20^{\circ}

k = 388 W/mK

h = 20\ W/m^{2}K

Now,

Perimeter of the fin, p = \pi D = 0.005\pi \ m

Cross-sectional area of the fin, A = \frac{\pi}{4}D^{2}

A = \frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}

To calculate the heat transfer rate:

Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})

where

m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237

Now,

Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W

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kondaur [170]

Answer:

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Explanation:

Given;

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v² = u² + 2as

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N is the normal reaction = mg

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