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
I'm not too sure what your asking but here are two answers that may help.
The ear drum amplifies the vibrations.
The cochlea changes vibrations into electric signals.
Answer:
Becomes greater
Explanation:
Distance between the central maxima and the next line increase with the wavelength i.e.
As, the wavelength of the red light is greater than the wavelength of the green light, thus, the distance between the central maxima and the next line becomes greater.
Option 5 is correct.
Answer:
16.6 kJ/°C
Explanation:
given,
Amount of heat absorbed = 45 kJ
initial temperature, T₁ = 25.5°C
final temperature, T₂ = 28.2°C
change in temperature = T₂ - T₁
= 28.2 - 25.5 = 2.7° C
Heat capacity of the object is equal to 16.6 kJ/°C
Work is calculated as the product of Force, Distance, and
angular motion. In this case, the work done by gravity is perpendicular to the motion
of the cart, so θ = 90°
and W=Fdcosθ
W=35.0 N x 20.0 m x cos90
W=0 J
This means that work done perpendicular to the direction of
the motion is always zero.
