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Nana76 [90]
3 years ago
9

Hey, guys need your help. I do school online and I have this project due. The problem is, I didn't know I needed most of these s

upplies till now. Can somebody please test this for me and give me a good explanation of what happened so I can understand what's going on? Thanks in advance! :)
Physics
1 answer:
Alexxandr [17]3 years ago
6 0
You need to post a picture so someone can help
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What is difference between heat engine and carnot engine
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Answer: A heat engine uses temperature differences which cause pressure changes to exert force on a moving part. A Carnot Process is a theoretical explanation of a process involving pressure and temperature changes during ,amongst other things, phase changes.

Explanation:

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How do gamma rays use electromagnetic waves to function....PLS HELP
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Gamma-rays have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum. These waves are generated by radioactive atoms and in nuclear explosions.

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Gamma-rays can kill living cells, a fact which medicine uses to its advantage, using gamma-rays to kill cancerous cells.

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What are the only elements that exist in nature as isolated atoms
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(that is helium, neon, argon, krypton, xenon and radon)

Term: Monoatomic

Explanation:

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Two asteroids are 75000 m apart one has a mass of 8 x 10^7 kg if the force of gravity between them is 1.14 what is the mass of t
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3 years ago
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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
hodyreva [135]

Answer : The correct option is, (c) 3.7\times 10^2J/^oC

Explanation :

First we have to calculate the energy or heat.

Formula used :

E=V\times I\times t

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

E=(3.6V)\times (2.6A)\times (350s)

E=3276J

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}

where,

C = heat capacity of the calorimeter

T_{initial} = initial temperature = 20.3^oC

T_{final} = final temperature = 29.1^oC

Now put all the given values in this formula, we get:

C=\frac{3276J}{(29.1-20.3)^oC}

C=372.27J/^oC=3.7\times 10^2J/^oC

Therefore, the heat capacity of the calorimeter is, 3.7\times 10^2J/^oC

7 0
3 years ago
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