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Vitek1552 [10]
3 years ago
6

A cook had a jar containing a sweet food and a jar containing a sour food. The sweet food has a strong attraction between its mo

lecules, and the sour food has a weak attraction between its molecules. At room temperature, both foods are liquids. The cook left both jars in a place where the same amount of energy was transferred into both substances. When she returned, one of the foods had changed phase while the other had not. Which food changed phase, and how did it change?
Physics
1 answer:
patriot [66]3 years ago
7 0

Answer:

The sweet food changed because the molecules were able to move fast enough to overcome the attraction between them with its molecules now moving away from each other.

Explanation:

We are told that the sweet food has a strong attraction between its molecules, and the sour food has a weak attraction between its molecules.

This means that the molecules in the sweet food would be moving at a faster rate than in the sour food because of the strong forces of attraction. Therefore, the molecules in the sweet food would be moving far away from each other hence the change of phase.

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The height increases (c)
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3 years ago
An overtone that's a whole number multiple of the fundamental frequency of a string is called a
alina1380 [7]
The correct answer in this case is B. Harmonic.

The remaining answers refer to other parts of a musical composition, and some such as the pitch can even be found in the study of human voice. 
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3 years ago
Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,
REY [17]

Answer:

a) 238U, 40K and 87Rb, b)   235U and to a lesser extent 40K , c)  he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0
3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

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viva [34]
The answer to this question is D or the last one 

4 0
3 years ago
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