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Evgesh-ka [11]
3 years ago
6

Why do you think the temperature does not change much during a phase change?

Physics
2 answers:
Zielflug [23.3K]3 years ago
8 0

Answer:

<em>By </em><em>the </em><em>there </em><em>is </em><em>no </em><em>change</em><em> </em><em>in </em><em>the </em><em>temperature</em><em> </em><em>until </em><em>the </em><em>phase </em><em>change </em><em>is </em><em>complete.</em><em> </em><em>It </em><em>is </em><em>tha</em><em>t</em><em> </em><em>during</em><em> </em><em>the </em><em>phase </em><em>change </em><em>the </em><em>energy </em><em>supplied </em><em>is </em><em>used </em><em>only </em><em>to </em><em>seperate </em><em>the </em><em>molecules </em><em>no </em><em>part </em><em>of </em><em>it </em><em>is </em><em>used </em><em>to </em><em>increase </em><em>the </em><em>relating </em><em>to </em><em>a </em><em>result </em><em>from </em><em>the </em><em>motion </em><em>energy</em><em> </em><em>of </em><em>the </em><em>molecules </em><em>so </em><em>it's </em><em>temperature</em><em> </em><em>will </em><em>not </em><em>rise </em><em>since </em><em>the </em><em>relating</em><em> to</em><em> a</em><em> </em><em>result</em><em> </em><em>from</em><em> the</em><em> </em><em>motion </em><em>energy </em><em>of </em><em>molecules </em><em>remains </em><em>the </em><em>same </em><em>.</em>

<em><u>I </u></em><em><u>hope </u></em><em><u>this </u></em><em><u>might </u></em><em><u>help </u></em><em><u>u</u></em><em> </em>

11111nata11111 [884]3 years ago
3 0

Answer:

It depends on where the temperature is dropping, in which body so to speak. Generally, the temperature adapts to the two bodies, for example if a hot piece of metal meets a cold one, the two will continue until they are at an equal temperature, an intermediate temperature.

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Vitek1552 [10]
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
                                                        mn=mass of neutron=<span>1.67 10^-27 kg
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Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5=  - 4.9 J
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the final answer is </span><span>Eb /nucleon </span><span>=  -5.625 x10^35 eV</span>
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The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
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V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
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Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

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