Th answer is A sorry if this isn’t what your looking for
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
Mass of yellow train, my = 100 kg
Initial Velocity of yellow train, = 8 m/s
mass of orange train = 200 kg
Initial Velocity of orange train = -1 m/s (since it moves opposite direction to the yellow train, we will put negative to show the opposite direction)
To calculate the initial momentum of both trains, we will use the principle of conservation of momentum which
The sum of initial momentum = the sum of final momentum
Since the question only wants the sum of initial momentum,
(100)(8) + (200)(-1) = 600 m/s
Lighting flows around the outside of a truck, and the majority of the current flows from the cars metal cage into the ground below. It's not very safe to be in a car or truck during bad weather.
