Answer:
hope this helps!
Explanation:
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K
The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
∴P1=1.103×105+40×103×9.8=493300Pa
We have T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=
Answer:
solution:
dT/dx =T2-T1/L
&
q_x = -k*(dT/dx)
<u>Case (1) </u>
dT/dx= (-20-50)/0.35==> -280 K/m
q_x =-50*(-280)*10^3==>14 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (3)
q_x =-50*(160)*10^3==>-8 kW
T2=T1+dT/dx*L=70+160*0.25==> 110° C
Case (4)
q_x =-50*(-80)*10^3==>4 kW
T1=T2-dT/dx*L=40+80*0.25==> 60° C
Case (5)
q_x =-50*(200)*10^3==>-10 kW
T1=T2-dT/dx*L=30-200*0.25==> -20° C
note:
all graph are attached
Answer:
Its color
Explanation: I got it right
Work = Force x Distance
Assuming that this work is being done parallel to the displacement that is, but under that assumption:
W = (50)(10)
W = 500 J
Answer:
The answer is first one 1...