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2 years ago

8. How much energy was expended by a rocked that is launched at 90 degrees

Physics

1 answer:

svlad2 [7]2 years ago

6 0

Answer:

because hsuehwbwnowbwbww9whw

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How is energy measured?

stepladder [879]

Because a Btu is so small, energy is usually measured in millions of Btus. 1 Btu = the amount of energy required to increase the temperature of one pound of water (which is equivalent to one pint) by one degree Fahrenheit. This is roughly the heat produced from burning one match.

<em>https://www.ucsusa.org/clean_energy/our-energy-choices/how-is-energy-measured.html</em>

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2 years ago

A stuntman jumps from the roof of a building to the safety net below. How far has the stuntman fallen after 1.6 seconds?

Fantom [35]

How tall is the building?
u need to divide the height by 1.6 to get your answer

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2 years ago

In the lesson a thermos is presented as an example of an isolated energy system. How could you change the thermos into an open e

iVinArrow [24]

If you want to change the thermos into an open energy system, you have to remove the lid. Once the lid is removed, the energy is no longer contained inside the thermos bottle. From the bottle, the energy dissipates to the environment.

7 0

3 years ago

Read 2 more answers

Define the relationship between force acceleration and mass

goblinko [34]

Force is mass times acceleration. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

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3 years ago

We are going to make an imaginary engine using water. We are going to heat 100 grams of water to 120 C from its initial temperat

Svetach [21]

Answer:

The work done by this engine is 800 cal

Explanation:

Given:

100 g of water

120°C final temperature

22°C initial temperature

30°C is the temperature of condensed steam

Cw = specific heat of water = 1 cal/g °C

Cg = specific heat of steam = 0.48 cal/g °C

Lw = latent heat of vaporization = 540 cal/g

Question: How much work can be done using this engine, W = ?

First, you need to calculate the heat that it is necessary to change water to steam:

$Q_{1} =m_{w} C_{w} (100-22)+m_{w}L_{w}+m_{w}C_{g}(120-100)$

Here, mw is the mass of water

$Q_{1} =(100*1*78)+(100*540)+(100*0.48*20)=62760cal$

Now, you need to calculate the heat released by the steam:

$Q_{2} =m_{w}C_{g}(120-100)+m_{w}L_{w}+m_{w}C_{w}(100-30)=(100*0.48*20)+(100*540)+(100*1*70)=61960cal$

The work done by this engine is the difference between both heats:

$W=Q_{1}-Q_{2}=62760-61960=800cal$

8 0

2 years ago