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Kobotan [32]
3 years ago
6

Which substance would require the most heat to melt 100 g of the substance?

Physics
1 answer:
IrinaVladis [17]3 years ago
3 0
Assuming all the substances are already at the temperature that is their melting point, we only need to worry about the heat required to change state, not the heat required to change temperature.

No calculations necessary for this question - just look at the latent heat of fusion. The higher this value, the more heat required per unit mass of the substance to melt it. Of the four answer options, aluminum has the highest value and therefore will take the most heat to melt the same mass.
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Select all that apply. which of the following astronomers supported the sun-centered system? tycho brahe johannes kepler coperni
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-Tycho Brahe and

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What is the scientific revolution??
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4 0
3 years ago
answers At the beginning of 2019, Robotics Inc. acquired a manufacturing facility for $12.3 million. $9.3 million of the purchas
pshichka [43]

Answer:

$613,077

Explanation:

Calculation for the depreciation on the building for 2021

First step is to calculate the depreciation expense for the year 2019 and 2020 using this formula

Depreciation expense = (Purchase price - Residual value) ÷ (Useful life)

Let plug in the formula

Depreciation expense= ($9,300,000 - $1,300,000) ÷ (20 years)

Depreciation expense= ($8,000,000) ÷ (20 years)

Depreciation expense= $400,000

Second step is to calculate the book value for the year 2021 using this formula

Book value = Purchase price - Depreciation expenses *2 years

Let plug in the formula

Book value= $9,300,000 - $400,000 × 2

Book value= $8,500,000

Last step is to calculate the depreciation for 2021 using this formula

2021 Depreciation=(Book value-Residual value)÷Useful life

2021 Depreciation= ($8,500,000 - $530,000) ÷ (15 years - 2 years)

2021 Depreciation=$7,970,000÷13 years

2021 Depreciation= $613,077

Therefore the depreciation on the building for 2021 will be $613,077

8 0
3 years ago
A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me
loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

Q = mC_p \Delta T

Where,

m = mass

C_p= Specific Heat

\Delta T = Change in temperature

Replacing with our values we have that

m = 0.1g

C_p = 139J/Kg\cdot K \rightarrow Specific heat of mercury

\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K

Replacing

Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J

Therefore the heat lost by mercury is 0.09J

5 0
3 years ago
A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 0.700 revoluti
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6 0
3 years ago
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