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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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100 points
luda_lava [24]

Answer:

10.21m/s²

Explanation:

Radial acceleration, also known as centripetal acceleration, refers to the acceleration of an object along the circular radius. It can be calculated using the formula:

a(r) = v²/r

Where a(r) = radial acceleration

v = velocity

r = radius of the circular path

According to the information provided in the question, velocity (v) = 17.5m/s, radius = 30m. Hence;

a(r) = v²/r

a(r) = 17.5²/30

a(r) = 10.208333

a(r) = 10.21m/s²

5 0
2 years ago
Light waves are electromagnetic waves that travel at 3.00 Light waves are electromagnetic waves that travel 108 m/s. The eye is
svlad2 [7]

(a) 5.45 \cdot 10^{14} Hz

The relationship between frequency and wavelength of an electromagnetic wave is given by

c=f \lambda

where

c=3.00 \cdot 10^8 m/s is the speed of light

f is the frequency

\lambda is the wavelength

In this problem, we are considering light with wavelength of

\lambda=5.50 \cdot 10^{-7} m

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/s}{5.50 \cdot 10^{-7} m}=5.45 \cdot 10^{14} Hz

(b) 1.83\cdot 10^{-15} s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

And using f=5.45 \cdot 10^{14} Hz as we found in the previous part, we can find the period of this wave:

T=\frac{1}{5.45 \cdot 10^{14} Hz}=1.83\cdot 10^{-15} s

5 0
3 years ago
Beyond what point must an object be squeezed for it to become a black hole
Naddik [55]

Answer:

Condition

\frac{m}{R} \geq \frac{c^{2} }{2*g}

Explanation:

Basically black hole is an object from which light rays can not escape it means   to go out from gravitational field , that body should thrown with speed greater then light.

Let's do some calculation

Gravitational potential at surface =-\frac{G*M*m}{R}

If we give kinetic energy equal to magnitude of Potential energy as on surface it will escape.

\frac{m*v^{2} }{2} =-\frac{G*M*m}{R}

⇒\frac{m}{R} =\frac{c^{2} }{2*G}

It will be more better for black hole if above ratio (analogous to density ) is more then above calculated

4 0
3 years ago
Which coordinate system would be most useful for two observers (one in Michigan and one in Florida) who wants to observe the sam
Mkey [24]

Answer : Celestial or azimuth - altitude

Explanation :  

Celestial : The celestial coordinates that are analogous to longitude and latitude are called RA and Dec.

RA = Right Ascension

Dec = Declination

RA is the measured in unit of time and Dec is measured in degree.

The equatorial coordinate system is the projection of the latitude and longitude coordinate system on the celestial sphere.

Azimuth - altitude :  Azimuth - altitude define the location of an object in the sky.

The altitude is the distance of an object appears to be above the horizon.

The Azimuth of the object is the angular distance  along the horizon to the location of the object.

3 0
2 years ago
Read 2 more answers
Is this answer correct?
alisha [4.7K]
Yes the answer is correct
4 0
3 years ago
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