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klio [65]
3 years ago
8

A flowerpot weighing 2 kilograms, what will be its speed at the end of 5 seconds?

Physics
1 answer:
Snezhnost [94]3 years ago
8 0

That depends on what its speed was at the beginning of 5 seconds,
and on what exactly happened to it.

If it got pushed off of the window sill and fell to the ground below,
then the answer would depend on how high the window sill is.

If the flower pot kept falling for the whole 5 seconds, then its speed
would be  <em>49 meters per second</em>  at the end of the 5 seconds.  That's
because here on Earth, gravity increases the speed of falling things
by 9.8 meters per second for every second they fall.

Naturally, this answer is only true if the window sill is more than
122.5 meters above the ground ... about 402 feet, or 40 floors.
If it's lower than that, then the flower pot hits the ground in less
than 5 seconds, and its speed at the end of 5 seconds is zero.

By the way ...

-- It makes no difference how much the flower pot weighs.
All objects fall with the same acceleration, and hit the ground
at the same time.

-- 5 kilograms is a mass, not a weight.
   

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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:

A.

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

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

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Q_1=-Q_2

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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

A.

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

B.

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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Answer:

B

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Explanation:

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