Answer:
The rate of transfer of heat is 0.119 W
Solution:
As per the question:
Diameter of the fin, D = 0.5 cm = 0.005 m
Length of the fin, l =30 cm = 0.3 m
Base temperature,
Air temperature,
k = 388 W/mK
h =
Now,
Perimeter of the fin, p =
Cross-sectional area of the fin, A =
A =
To calculate the heat transfer rate:
where
Now,
Answer:
A) The speed of the water must be 8.30 m/s.
B) Total kinetic energy created by this maneuver is 70.12 Joules.
Explanation:
A) Mass of squid with water = 6.50 kg
Mass of water in squid cavuty = 1.55 kg
Mass of squid =
Velocity achieved by squid =
Momentum gained by squid =
Mass of water =
Velocity by which water was released by squid =
Momentum gained by water but in opposite direction =
P = P'
B) Kinetic energy does the squid create by this maneuver:
Kinetic energy of squid = K.E =
Kinetic energy of water = K.E' =
Total kinetic energy created by this maneuver:
If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.
Answer:
170N
Explanation:
First add 530N to 150N and you get 680N, then add 400N to 450N and get 850N. So subtract 850N by 680N and you get 170N