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IRINA_888 [86]
3 years ago
14

Investigators are working on a case where they need to know whether a watch will stop when it is dropped. In order to have a ver

ified or reliable answer to this question, what should they do?
Call the manufacturer to ask whether it will stop under these circumstances
Drop a similar watch multiple times and record the results
Consult with more senior forensic scientists for their experiences
Take the watch apart and test the mechanism to see what causes failure
Physics
2 answers:
ivanzaharov [21]3 years ago
6 0

In order to have a verified or reliable answer to this question, they should take the watch apart and test the mechanism to see what causes failure.

Answer: Option D

<u>Explanation: </u>

For dropping a similar watch to record results, we need to buy another one. So, it becomes expensive. And, calling the manufacturer will give accurate results, but no knowledge. It is like getting multiple choices by luck.

Consulting senior faculty won’t result in any progress, because they may not have conducted these. So, one should rip off the watch, and study the mechanism, and its limitations.  The testing of mechanism gives the actual results.

vredina [299]3 years ago
5 0

Answer: B) Drop a similar watch multiple times and record the results.

Explanation:

In order to have a verified or reliable answer to this question, investigators must drop a similar watch multiple times and record the results. By doing this and repeatedly getting the same result(s), they are able to determine that their answer is in fact true.

***<em> I would also like to note that I got this question right on a test. Cheers!</em>

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vichka [17]

Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

Q_{net,in} - W_{net, out} = \Delta U (1)

Where:

Q_{net, in} - Net input heat, measured in joules.

W_{net, out} - Net output work, measured in joules.

\Delta U - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then \Delta U < 0, which is caused by three scenarios:

(i) Q_{net,in} < 0, W_{net, out} < 0, |Q_{net,in}|>|W_{net,out}|

(ii) Q_{net, in} > 0, W_{net,out} > 0, |Q_{net,in}|

(iii) Q_{net,in}< 0, W_{net, out}>0

In the case Q_{net,in} > 0, W_{net, out}, the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

Q_{net, in} = \Delta U, Q_{net, in} < 0 (2)

If Q_{net, in} < 0, then  \Delta U < 0. Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

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2 years ago
Q¹=0,07Mc Q²=2C r=1,08 cm F=.......?
Nostrana [21]
The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2)  where:
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</span>r is the distance between the two charges.

Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span> 
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Answer:

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