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:
D
Explanation:
Let’s calculate the kinetic energy for all of the choices.
a. (1/2)(100)(100)^2 = 50(10000)=500,000
b. (1/2)(100)(1)^2 = 50
c. (1/2)(10)(100)^2 = 5(10000) = 50,000
d. (1/2)(1)(1)^2 = 0.5
We can see that (d) has the least kinetic energy.
Answer:
Acceleration, 
Explanation:
It is given that,
Separation between the protons, 
Charge on protons, 
Mass of protons, 
We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :


So, the acceleration of two isolated protons is
. Hence, this is the required solution.
Answer:
The horizontal range will be 
Explanation:
We have given initial speed of the shell u = 
Angle of projection = 51°
Acceleration due to gravity 
We have to find maximum range
Horizontal range in projectile motion is given by

So the horizontal range will be 
Answer:
C
Explanation:
Because this same question was on my test last week and I got it correct