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aalyn [17]
1 year ago
8

A block is projected with speed v across a horizontal surface and slides to a stop due to friction. The same block is then proje

cted with the same speed v up an incline where is slides to a stop due to friction. In which case did the total mechanical energy of the block decrease the least
Physics
1 answer:
natka813 [3]1 year ago
8 0

Answer: C. The case on the inclined surface had the least decrease intotal mechanical energy.

Explanation:

First and foremost, it should be noted that the mechanical energy is the addition of the potential and the kinetic energy.

From the information given, it should be known that when the block is projected with the same speed v up an incline where is slides to a stop due to friction, the box will lose its kinetic energy but there'll be na increase in the potential energy as a result of the veritcal height. This then brings about an increase in the mechanical energy.

Therefore, the total mechanical energy of the block will decrease the least when the case on the inclined surface had the least decrease intotal mechanical energy.

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Answer:

\eta=0.5074\ or\ 50.74\%

Explanation:

<em><u>Considering the density & specific heat capacity of coffee to be equal to that of water.</u></em>

<em><u>GIVEN:</u></em>

  • density \rho=1\ g.mL^{-1}
  • specific heat c=4.186\ J.g^{-1}.K^{-1}
  • mass of coffee, m=200\times 1=200\ g
  • initial temperature of coffee, T_i=30^{\circ}C
  • final temperature of coffee, T_f=60^{\circ}C
  • power rating of oven, P=1100\ W
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<u>Heat released by the coffee to come to 60°C:</u>

Q=m.c.\Delta T

Q=200\times 4.186\times 30

Q=[tex]\eta=\frac{25116}{49500}\ J[/tex]

<u>Now the energy used by the oven in the given time:</u>

E=P.t

E=1100\times 45

E=49500\ J

Now the efficiency:

\eta=\frac{Q}{E}

\eta=0.5074\ or\ 50.74\%

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