Answer:
Δr=20.45 %
Explanation:
Given that
Rake angle α = 15°
coefficient of friction ,μ = 0.15
The friction angle β
tanβ = μ
tanβ = 0.15
β=8.83°
2φ + β - α = 90°
φ=Shear angle
2φ + 8.833° - 15° = 90°
φ = 48.08°
Chip thickness r given as
r=0.88
New coefficient of friction ,μ' = 0.3
tanβ' = μ'
tanβ' = 0.3
β'=16.69°
2φ' + β' - α = 90°
φ'=Shear angle
2φ' + 16.69° - 25° = 90°
φ' = 49.15°
Chip thickness r' given as
r'=0.70
Percentage change
Δr=20.45 %
Answer: Aerosol Cans
Explanation: I just did the quiz
Question:
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.
For a power line that supplies power to 10 000 households, we can conclude that
a) IV < I²R
b) I²R = 0
c) IV = I²R
d) IV > I²R
e) I = V/R
Answer:
d) IV > I²R
Explanation:
In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.
The power delivered to households is given by
P = IV
The losses in the transmission line are given by
Ploss = I²R
Therefore, the relation IV > I²R holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.
Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.
