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labwork [276]
3 years ago
11

Help!! HELP HELP HELP HELP

Physics
1 answer:
Anarel [89]3 years ago
8 0

Answer:

if a red <u>gaint </u> star is large enough, it may eventually blow up in an explosion called a <u>supernova</u><u> </u> leaving behind very dense <u>neutron</u> star.

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If an object has a large momentum, how easy is it to stop, and why?
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Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a
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Answer:

Approximately 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}} assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity c and mass m. Increasing the temperature of this object by \Delta T would require Q = c\, m \, \Delta T.

Look up the specific heat of water: c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}.

It is given that the mass of the water in this mixture is m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}.

Temperature change of the water: \Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}.

Thus, the water in this mixture would have absorbed :

\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}.

Thus, the energy that water absorbed was: Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}.

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, Q(\text{water}), would be the opposite of the energy that the metal in this mixture released.

Thus: Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J} (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}.

Temperature change of the metal in this mixture: \Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}.

Rearrange the equation Q = c\, m \, \Delta T to obtain an expression for the specific heat capacity: c = Q / (m\, \Delta T). The (average) specific heat capacity of the metal pieces in this mixture would be:

\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}.

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3 years ago
Balanced forces keep this boy in place on his chair. What
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While the boy is sitting on the chair it creates a force downward on the chair and therefore the chair takes it and gives off the equal amount of force. So while he is putting force downward the chair is putting the same force upward.

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