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:
4960 N
Explanation:
First, find the acceleration.
Given:
v₀ = 6.33 m/s
v = 2.38 m/s
Δx = 4.20 m
Find: a
v² = v₀² + 2aΔx
(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)
a = -4.10 m/s²
Next, find the force.
F = ma
F = (1210 kg) (-4.10 m/s²)
F = -4960 N
The magnitude of the force is 4960 N.
We know that the change in momentum is equals to the product of force and time that is impulse ( 
). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,
Here, u is initial velocity which is zero.
.
Thus, impulse
From Newton`s second law,
Therefore, impulse
Given, 
and 
Substituting these values, we get
Change in momentum = impulse
.
The standard wave format for any wave is transverse wave
Answer:
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