Answer:
The objects kinetic energy increases as it falls from some height.
Answer:
Explanation:
Given the following ;
- TA = 77°C and TB = 62°C at x1, KA = 180w/mk
- Tb = 100°C and T[infinity] = 25°C
The two rods could be approximated as a fins of infinite length.
- TA = 77 0C, θA = (TA - T∞) = 77 - 25 = 52 0C
- TB = 62 0C , θB = (TB - T∞) = 62 - 25 = 37 0C
- Tb = 100 0C , θb = (Tb - T∞) = (100 - 25) = 75 0C
The temperature distribution for the infinite fins are given by ;
- θA/θb= e-√(hp/A.kA) x1 ....................(1)
- θB/θb = e-√(hp/A.kB) x1.......................(2)
Taking natural log on both sides we get,
- In(θA/θb) = -√(hp/A.kA) x1 ...................(3)
- In(θB/θb) = -√(hp/A.kB) x1 .....................(4)
- [ In(θA/θb) / In(θB/θb)] = √(KB/KA)
- [ In(52/75) / In(37/75)] = √(KB/180)
- KB = 48.36W/m. K, Hence The thermal conductivity of the second material is KB = 48.36 W/m.K
Answer: 6.64 × 10^ -6 m
Explanation:
for a constructive interference;
ρ= 2πm
remember that we are looking for the seventh bright side fringe.
so, m=7
ρ=14π
We now have to find the phase different
ρ`=2πx/ρ - 2πx/ρ
p` is the wavelenght in the mica. x= width of the mica wavelength in the mica.
ρ= ρ/n.
n= refractive index.
The phase will now be
ρ`= 2πx/ρ - 2πx = 2πx(n-1)/ρ.
ρ= ρ`
14π = 2πx (n-1)/ ρ
we now solve for x; x = 7ρ/n-1
x = 7(550× 20^ 9)/ 1.58 - 1
= 6.64 × 10^-6
