Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:
diameter = 40 mm ==> area = <em>π</em> (0.04 m)² ≈ 0.00503 m²
diameter = 120 mm ==> area = <em>π</em> (0.12 m)² ≈ 0.0452 m²
Then the speed at the end of the 40 mm pipe is
(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s
(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s
Answer: Thermal comductivity (K) is 3.964x 10 ^-3 W/m.k
Explanation:
Thermal comductivity K = QL/A∆T
Q= Amount of heat transferred through the material in watts = 75W
L= Distance between two isothermal planes = 0.740mm
A= Area of the surface in square metres = 2m^2
∆T= Temperature change = (37-30) °C.
Solving this : K =( 75 x 0.740 x 10^-3)/ 2 x (37-30)
K = 3.964x 10 ^-3 W/m.k
Answer:
Explanation:
When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.
When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other .So both the shells lose their charges .The positive half shell also loses all its charges
When we separate the half shells , there will be no deflection in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.
the first one is medium, the second one is type, and the third one is temperature
. if i gave the correct answer, please give best answer x
Say the initial point is (0,0)
The final point is
x = 200 + 135*cos(30) = 200 + 135*sqrt(3)/2 = 316.91 ft
y = 135*sin(30) = 135/2 = 67.5 ft
Resultant vector = (316.91, 67.5) - (0,0) = 316.91, 67.5) ft
