Answer:
(a) W = 1329.5 J = 1.33 KJ
(b) ΔU = 24.27 KJ
Explanation:
(a)
Work done by the gas can be found by the following formula:
where,
W = Work = ?
P = constant pressure = (0.991 atm)(
) = 100413 Pa
ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)(
) = 0.01324 m³
Therefore,
W = (100413 Pa)(0.01324 m³)
<u>W = 1329.5 J = 1.33 KJ</u>
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(b)
Using the first law of thermodynamics:
ΔU = ΔQ - W (negative W for the work done by the system)
where,
ΔU = change in internal energy of the gas = ?
ΔQ = heat added to the system = 25.6 KJ
Therefore,
ΔU = 25.6 KJ - 1.33 KJ
<u>ΔU = 24.27 KJ</u>
no cause kids will think its Neglect and names will hert the kids and think they are not wanted
Picosecond = 10 ^ -12 seconds.
Zeptosecond = 10^ -18 seconsds
Petaseonds = 10^15 seconds
To express Picoseconds into any of other two, you have to divide 10^-12 by the power index of the one in question
1Picosecond : 10^-12 / 10^-18 = 10^ (-12- 18) = 10^ (-12+18)= 10^6 zeptoseconds
1Picosecond : 10^-12 / 10^15 = 10^ (-12-15) = 10^-27 Petaseconds.
1Picosecond = 10^6 zeptoseconds
1Picosecond = 10^-27 Petaseconds
Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
