Answer:
k=0.0577w/mk
k=0.03338BTU/h-ft-ºF
Explanation:
First we use the equation for heat transfer by conduction and solve for K
Q=KA(T2-T1)/x
Where
q=heat
A=area
T=temperature
K=conductivity
x=thickness
solving for k
k=Qx/A(T2-T1)
Q/A=Heat flux=q
k=qx/(T2-T1)
k=35.1*0.025/(318.4-303.2)
K=0.0577w/mk
remember that 1 btu / hftk equals 1,729 w / mk
K=0.0577/1.729=0.03338BTU/h-ft-ºF
U = 6.5 m/s, initial speed
t = 3.6 s, time
a = 0.92 m/s², acceleration
Let v = the final velocity.
Then
v = u +at
v = (6.5 m/s) + (0.92 m/s²)*(3.6 s) = 9.812 m/s
Answer: 9.81 m/s
Answer:As the size of a star increases, luminosity increases. If you think about it, a larger star has more surface area. That increased surface area allows more light and energy to be given off. Temperature also affects a star's luminosity.
Answer:
how ever many layers it has i think
Explanation:
