Answer:
340 m/s
Solution:
As per the question;
Time, t = 3.0 s
Total distance moved by the ship, d = 10,500 km
The increase in speed, v = 340 m/s
Now,
To calculate the U.S.S Burger's change in speed, 
:
The final velocity is given by:
(1)
where
Also, the change in velocity is given by:
(2)
Now, from eqn (1) and (2):
Heated air rises while cooled air sinks.
Answer:
There are 6 electrons in the outermost shell.
Explanation:
Sulphur is a non-mettalic element which is in the period 3 and group .6on the periodic table. It has an atomic number of 16 and a Mass number of 32. Atomic number tells you the number of electrons in an electrically neutral atom. It has the electronic configuration of 1s2 2s2 2p6 3s2 3p4.
The orbitals have a formula 2n^2 where n = 0, 1, 2, 3 etc.
In the shells, n = 1 so there are 2 electrons. For n = 2, 2*(2)^2 = 8 electrons. So, 16 - (8 + 2) = 6 electrons in the 3 shell (outermost shell)
Therefore from the electronic confriguration above, there are 6 electrons in the outermost shell.
Answer:
Mulitple Changes.
Explanation:
Most people want something to be done, and never change again. But some others do, since they have there own beliefs. Alot of people are concerned about tax rates, taxes, nationial polices, own beliefs. People have there own beliefs and when someone is elected into the federal Government, they have there own Beliefs (Mostly Republican Or Democrate). People go against on what they dont think and go with what they think. At the end whoever sways the people on who to believe and vote for, Is elected.
Answer:
The thermal conductivity of the wall = 40W/m.C
h = 10 W/m^2.C
Explanation:
The heat conduction equation is given by:
d^2T/ dx^2 + egen/ K = 0
The thermal conductivity of the wall can be calculated using:
K = egen/ 2a = 800/2×10
K = 800/20 = 40W/m.C
Applying energy balance at the wall surface
"qL = "qconv
-K = (dT/dx)L = h (TL - Tinfinity)
The convention heat transfer coefficient will be:
h = -k × (-2aL)/ (TL - Tinfinty)
h = ( 2× 40 × 10 × 0.05) / (30-26)
h = 40/4 = 10W/m^2.C
From the given temperature distribution
t(x) = 10 (L^2-X^2) + 30 = 30°
T(L) = ( L^2- L^2) + 30 = 30°
dT/ dx = -2aL
d^2T/ dx^2 = - 2a
