The valence electrons are the one furthest from the nucleus
Answer:
a) 90 kJ
b) 230.26 kJ
Explanation:
The pressure at the first point = 10 bar —> 10 x 102 = 1020 kPa
The volume at the first point = 0.1 m^3
The pressure at the second point = 1 bar —> 1 x 102 = 102 kPa
The volume at the second point = 1 m^3
Process A.
constant volume V = C from point (1) to P = 10 bar.
Constant pressure P = C to the point (2).
Process B.
The relation of the process is PV = C
Required
For process A & B
(a) Sketch the process on P-V coordinates
(b) Evaluate the work W in kJ.
Assumption
Quasi-equilibrium process
Kinetic and potential effect can be ignored.
Solution
For process A.
V=C
There is no change in volume then
The work is defined by
║ V║limit 1--0.1
90 kJ
Process B
PV=C
By substituting with point (1) C = 10^2 x 1= 10^2
The work is defined by
║ ln(V)║limit 1--0.1
=230.26 kJ
Answer:
both experience forces or at least a force
Explanation:
it would go in the direction the other object
(second object, the one that crashed) was going
si if going right then right if left then left
plus or minus
Answer:
= 38.89 m/s
Explanation:
speed of first car (u) = 80 km/h = 22.22 m/s
time (t) = 30 s
distance (d) = 500 m
find the speed of second car (v)
let the speed of the second car be represented as 'v'
We can get the speed of the second car from the formula net speed = distance between cars / time interval
where
- distance between cars = 500 m
- time interval = 30 s
- net speed = v-22.22 (since both cars move in the same direction)
- substituting the above into the formula we have
v-22.22 = 500 / 30
v = 500/30 + 22.22
v = 38.89 m/s