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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:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

F = -kx

And Newton's Second Law also states that

F = ma = m\frac{d^2x}{dt^2}

Combining two equations yields

a = -\frac{k}{m}x

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

\omega = \sqrt{\frac{k}{m}}

And given that ω = 2πf

the relation between frequency and mass becomes

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}.

Let's apply this to the variables in the question.

0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg

6 0
3 years ago
A baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg. What’s the baseballs velocity?
Mashcka [7]

Answer:

<em>v=40 m/s south</em>

Explanation:

<u>Momentum </u>

It's a physical magnitude that measures the product of the mass by the velocity of a particle. Its units in the International System is kg.m/s and the formula is

p=m.v

Where m is the mass and v the velocity of the particle. If we wanted to solve for v, we have

\displaystyle v=\frac{p}{m}

The baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg, thus

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. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of
Rina8888 [55]

The force required is 319 N

Explanation:

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F_f = \mu W

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Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

\mu = 0.55 (coefficient of friction)

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Substituting, we find:

F=(0.55)(580)=319 N

Learn more about force of friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

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