4. Static, sliding,rolling,and fluid friction
Answer:
Hello, there!
In order to solve problems like this DO PEMDAS
The Exponent:
So, equals 4
Your problem becomes
Now we have to do what is in the (
( the is a )
And
Problem becomes
So, this means that your answer would be
Good luck on your assignment and enjoy your day!
kjExplanation:
From the equation q=mCΔT, set the q of copper = to q of water,
So --- mCΔT(copper)=mCΔT(water).
mass (Cu - copper) = 38g
mass (H2O - water) = 15g
C (H2O) = 4.184 J/g*C
ΔΤ (H2O) = 33-22 = 11*C
ΔΤ (Cu) = 33-80 = -47*C (the final temp is the same for both materials - thermal equilibrium)
C (Cu) = ?
So --- 38(-47)C[Cu]=15(4.184)(11)
--- C[Cu]=690.36/(-1786) = 0.3865 J/g*C, or 0.39 in 2 sig figs. (The negative goes away, because specific heats are usually positive)
The energy containing electron transporters of FADH2 are not produced during glycolysis.
Answer:
See pictures!
Explanation:
Here is a drawing of butane (C4H10); labeled #1. An isomer of butane will also have the chemical formula C4H10 but is depicted different geometrically. I labeled this one #2.
