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3 years ago

How much of the total energy in Problem 3 and 4 has been transformed to kinetic energy?

Physics

1 answer:

ale4655 [162]3 years ago

6 0

I am only a HS Freshman so I might be wrong

Answer:

3. Its potential energy and the formula for PE = mgh so you would use the weight given (Weight = mg) and mg is in the (Potential Energy) forumula so we multiply 3 by 30 and get 90 Joules

4. We would do the same thing with this: PE = mgh; PE = 10*3 = 30 Joules

Once again I am only a Freshman and I have taken Private AP Physics classes so just maybe wait for someone who is more confident.

This is just what I would answer

btw, for problem 5, the fromula for kinetic energy is 1/2*m*v^2

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En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

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Answer:

c=0.14J/gC

Explanation:

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

We can use the expression for heat transmission

$Q=mc(T_2-T_1)$

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

$Q_1=-Q_2$

for water we have to

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replacing we have

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

I hope this is useful for you

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

Podemos usar la expresión para la transmisión de calor

$Q=mc(T_2-T_1)$

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

$Q_1=-Q_2$

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

$c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}$

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