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Rudiy27
3 years ago
5

Air in a cylinder is compressed to one-tenth of its original volume without change in temperature. What happens to its pressure?

Imagine now that a valve is opened in order to restore the initial pressure value. What percentage of the molecules have escaped?
Physics
1 answer:
alexandr402 [8]3 years ago
3 0
<h2>One-tenth of Original Volume</h2>

Explanation:

  • Air in a cylinder is compressed to one-tenth of its original volume without  a change in temperature then the pressure will be increased
  • If the valve is opened in order to restore the initial pressure value the molecules will be escaped
  • The percentage of the molecules that have escaped to attain the initial pressure will be equal to the one-tenth of its original volume
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when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv
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3 years ago
States that there is an exchange of materials when two objects come into contact with each other
pishuonlain [190]

"Edmond Locard" states that there is an exchange of materials when two objects come into contact with each other.

<u>Explanation:</u>

A French criminologist who was popular as the "Sherlock Holmes of France," the pioneer in forensic science named as Dr. Edmond Locard. He articulated forensic science's fundamental principle "Each touch leaves a trace." This became known as Locard's philosophy of exchange. A Locard hypothesized that each and every time you touch another person, place or object, the result would be an exchange of materials. Burglars, for instance, will leave evidence of their existence behind and take traces with them too.

3 0
3 years ago
Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t
lorasvet [3.4K]

Answer:

V_3\approx 4.28\,\,V

I_1=0.0572\,\,amps

I_3\approx 0.171\,\,amps

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega

By knowing this, we can estimate the total current through the circuit,:

Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps

So approximately 0.17  amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V

So now we know that the potential drop across the parellel resistors must be:

10 V -  4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps

4 0
3 years ago
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Ymorist [56]

Answer:

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L' = (2*0.5) / (0.5 + 1.8)

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5 0
3 years ago
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