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
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
Formula for terminal
velocity is:
Vt = √(2mg/ρACd)
<span>Vt = terminal velocity = ?
<span>m = mass of the falling object = 72 kg
<span>g = gravitational acceleration = 9.81 m/s^2
<span>Cd = drag coefficient = 0.80
<span>ρ = density of the fluid/gas = 1.2 kg/m^3</span>
<span>A = projected area of the object (feet first) = 0.21 m * 0.41
m = 0.0861 m^2
Therefore:</span></span></span></span></span>
Vt = √(2 * 72
* 9.81 / 1.2 * 0.0861 * 0.80)
<span>Vt = 130.73 m/s</span>
26) A) 400J as GPE = mgh
27) C) Both obj has same PE
28) B) It decreases as energy is not being provided.
29) C) (42 - 0)/7 = 6m/s2
Hello,
The answer should be option D "specific heat".
Reason:
Specific heat is what tells the person how my heat and or pressure is required to raised the objects temperature. Its not option A convection because I'm convection is mainly used as like the ozone layer and the suns rays. Its not option B radiation because radiation doesn't tell the person what temperature it needs in order for it to rise its more like its already rises and hot. Its also not option C because conduction is the process of a object heating another object therefore the answer is option D.
If you need anymore help feel free to ask me!
Hope this helps!
~Nonportrit
