Answer:
<h2>698.3Kpa</h2>
Explanation:
Step one:
given data
V1=0.25m^3
T1=290k
P1=100kPa
V2=0.5m^2
T2=405k
P2=? final pressure
Step two:
The combined gas equation is given as
P1V1/T1=P2V2/T2
Substituting we have
(100*0.25)/290=P2*0.05/405
25/290=0.5P2/405
0.086=0.05P2/405
cross multiply
0.086*405=0.05P2
34.9=0.05P2
divide both sides by 0.05
P2=34.9/0.05
P2=698.3Kpa
<u>Therefore the new pressure is 698.3Kpa when the gas is compressed</u>
Answer:Vb=-6i-(-0.1ωab+8)j m/s
Explanation:
Va=V0+Va0
Va=V0+(ra0 x ωao)
ω=Angular velocity of link A0
Using r0a=0.1m;
Va=V0+(0.1i x ω0a K)
Va=0
ixk=j
Va=0+0.1ω0aj
Calculating te velocity of using te equation below
Vb=Va+Vba
Vb=Va+ωab x rba
ωab=40rad/s
rab=-0.21i+0.15j
Va=0.1ω0aj
Vb=Va+ωabxrba
Vb=0.1ω0aj+40k x -(0.21i+0.15j)
Vb=0.1ω0aj-8j-6i
Vb=-6i-(-0.1ωab+8)j m/s
Answer:
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Answer:i think it is 35
Explanation:
i just guessed sorry im only in 5th grade
