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nadezda [96]
3 years ago
5

A volleyball is spiked so that its incoming velocity of +2.63 m/s is changed to an outgoing velocity of -20.2 m/s. The mass of t

he volleyball is 0.350 kg. What is the magnitude of the impulse that the player applies to the ball?
Physics
1 answer:
earnstyle [38]3 years ago
8 0

Answer:

The impuise is 7.9905 kg*m/s

Explanation:

Step one:

given data

v1= +2.63m/s

v2=-20.2m/s

mass m= 0.350kg

Step two:

From the expression for impulse

Ft= mΔv

substituting our data into the expression we have

Ft= 0.35*(-20.2-2.63)

Ft= 0.35*22.83

Ft=7.9905 kg*m/s

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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

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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

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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

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

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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:

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Explanation:

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