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:by driving east 3 blocks from the starting point
Explanation:)
Answer:
initial magnetic field 1.306 T
Explanation:
We have given area of the conducting loop 
Emf induced = 1.2 volt
Initial magnetic field B = 0.3 T
Time dt = 0.087 sec
We know that induced emf is given by 


So initial magnetic field = 1.606-0.3= 1.306 T
It doesn't matter what the object's initial velocity is, or how long
the acceleration lasts. All that matters is the object's mass and
acceleration.
Force = (mass) x (acceleration) =
(5kg) x (15 m/s²) =
75 kg-m/s² = <em>75 newtons .</em>