Answer:
Explanation:
We need the power equation for this which is
P = Work/time
We have everything we need to solve this (the mass of the object is extra information):
P = 6860/4
P = 1715W
<h2><em>THERE WAS A TORNADO????????????</em></h2><h2><em></em></h2><h2><em>SAY WHAT?!</em></h2>
- 9.5 m/s^2
use the SUVAT method
0 = 38 + 4a
Answer:
<em>1.01 W/m</em>
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.
area of the pipe per unit length A = = m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>