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2 years ago

7

What’s the equation that links with total input, efficiency, energy and useful output energy transfer

1 answer:

weqwewe [10]2 years ago

8 0

Efficiency. The ratio of energy which was transferred to a useful form compared to the total energy initially supplied is called the efficiency of the device. Efficiencies can be written as decimals like 0.33 or percentages 33%. To convert a efficiency expressed as a decimal to a percentage you need to multiply by 100.

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A blue puck has a velocity of 3i –4j m/s. Its mass is 20 kg. What is its momentum?

damaskus [11]

P = m * v
v = {3i - 4j} = square root (3^2 + 4^2) = 5
P = 20 * 5
P = 100 kg m/s

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A car hits another and the two bumpers lock together during the collision. is this an elastic or inelastic collision?

valkas [14]

Inelastic.
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Pavlova-9 [17]

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2 years ago

A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa

gavmur [86]

Answer : The correct option is, (d) $535^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of copper = $0.10cal/g^oC$

$c_2$ = specific heat of water = $1.00cal/g^oC$

$m_1$ = mass of copper = 120 g

$m_2$ = mass of water = 300 g

$T_f$ = final temperature of mixture = $35^oC$

$T_1$ = initial temperature of copper = ?

$T_2$ = initial temperature of water = $15^oC$

Now put all the given values in the above formula, we get:

$120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC$

$T_1=535^oC$

Therefore, the temperature of the kiln was, $535^oC$

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2 years ago

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Sonja [21]

I am pretty sure the answer would be too stretch

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