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Paraphin [41]
3 years ago
6

Kinetic energy is the energy of ______. It increases with mass ______ and _______ .Potential energy is the energy of _______or__

________- . It increases with ________, and ______.
Physics
1 answer:
Alona [7]3 years ago
5 0

Answer:

velocity.  height. weight. possition. place. energy. force.

Explanation: 50/50 % chance they are wrong and write.

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Why do organs have different types of tissues?<br><br><br> PLZ HELP
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its hard to explain its very complex but its so they can function properly

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Match each situation
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A: is potential
C: is losing kinetic energy and gaining potential energy
B: kinetic energy is at its highest
D: is loosing potential energy and gaining kinetic energy
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Based upon the information you have learned throughout this module on blood spatter analysis, do you feel that analyzing blood s
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I do not think that it is the most reliable way to gain information since it is very hard to do and can be easily messed up. No, I don't think you can charge someone on only evidence from blood spatter, but if there was additional evidence I think that this would definitely help with the case but not on its own, since it doesn’t give you physical evidence about the suspect.

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A car moving around a circular track with constant speed (Select all that apply.) has an acceleration with a direction that cann
BARSIC [14]

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5 0
3 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
4 years ago
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