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
a) create an expression
for the ball's initial horizontal velocity, V0x, in terms of the variables
given in the problem statement.
v0x = vf * cos(Θf)
<span>
b) calculate the ball's initial vertical velocity, V0y, in
m/s</span>
v0x = 32.4m/s * cos(-25.5º)
= 29.2 m/s <span>
tanΘ = v1y / v0x → tan(-25.5) = v1y / 29.2m/s → v1y = -13.93
m/s
the vertical velocity when the ball was caught.
(v0y)² = (v1y)² + 2as = (-13.93m/s)² + 2 * 9.8m/s² * 5.5m = 301.78
m²/s²
v0y = 17.37 m/s
c) calculate the magnitude of the ball's initial velocity,
v0, in m/s</span>
v0 = sqrt (v0y^2 +
v0x^2)
v0 = sqrt (17.37^2 + 29.2^2)
m/s
v0 = 33.98 m/s
<span>
d) find the angle, theta0, in degrees above the horizontal at
which which the ball left the bat.</span>
tan Θ = v0y/v0x
<span>Θ = arctan(17.37/29.2) =
30.75º above horizontal</span>
Answer:
Explanation:
As we know that loop is placed in YZ plane and magnetic field is along x direction
So here net force on the side of the loop which lies along Y axis is given as
here we know that on Y axis z = 0
so B = 0
so we have
now on the opposite side we have z = a
so magnetic field is given as
so force on that side is given as
so net force on the loop is given as
In which of the Earth's layers are diamonds formed? -
Diamonds form in the Earth's mantle, a thick layer between the thin crust and Earth's metal core.
<span>Answer:
Using 1/f = 1/d' + 1/d ...(where d' object distance and d is image distance)
1/4 = 1/7 + 1/d
1/4 - 1/7 = 1/d
3/28 = 1/d
d = 28/3
d = 9.33 cm</span>