Answer:
Given: m = 5 kg, r =1 m, ω = 2 radian/s, centripetal force = ?
Formula : Centripetal force = mv²/r = mrω²
(Since v = r ω)
Put the values.
Answer:
A) P1=2 [bar] , W=-12 [kJ]
B) P1=0.8 [bar] , W=-7.3303 [kJ]
C) P1=0.6077 [bar] , W=-6.4091 [kJ]
Explanation:
First, from the problem we know the following information:
V1=0.1 m^3
V2=0.04 m^3
P2=2 bar =200 kPa
The relation PV^n=constant means PV^n is a constant through all the process, so we can derive the initial pressure as:
a) To the case a) the constant n is equal to 0, we can calculate the initial pressure substituting n=0 in the previous expression, so:
The expression to calculate the work is:
If n=0:
Then:
The work is:
b) To the case b) the constant n is equal to 1, we can calculate the initial pressure substituting n=1 in the initial expression, so:
If n=1 then:
To calculate the work:
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Substituting:
c) To the case c) the constant n is equal to 1.3, we can calculate the initial pressure substituting n=1.3 in the initial expression, so:
First:
The work:
Substituting:
W=-6.4091 kJ
Answer:
v = 1.36 cm / y
Explanation:
For this exercise we must assume that the displacement of the plates is constant over time, so we will use the kinematic relationships for the uniform movement
v = d / t
We reduce the quantities to the SI system
d = 320 km (1000 m / 1km) (100 cm / 1 m)
d = 3.2 107 cm
let's calculate
v = 32.107 / 23.5 106
v = 1.36 cm / y
As water vapour cools, it changes back into liquid form. drops come together to form 'clouds'? maybe