Answer:
A) 60%
B) p2 = 1237.2 kPa
v2 = 0.348 m^3
C) w1-2 = w3-4 = 1615.5 kJ
Q2-3 = 60 kJ
Explanation:
A) calculate thermal efficiency
Л = 1 -
where Tl = 300 k
Th = 750 k
hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%
B) calculate the pressure and volume at the beginning of the isothermal expansion
calculate pressure ( P2 ) :
= P3v3 = mRT3 ----- (1)
v3 = 0.4m , mR = 2* 0.287, T3 = 750
hence P3 = 1076.25
next equation to determine P2
Qex = p3v3 ln( p2/p3 )
60 = 1076.25 * 0.4 ln(p2/p3)
hence ; P2 = 1237.2 kpa
calculate volume ( V2 )
p2v2 = p3v3
v2 = p3v3 / p2
= (1076.25 * 0.4 ) / 1237.2
= 0.348 m^3
C) calculate the work and heat transfer for each four processes
work :
W1-2 = mCv( T2 - T1 )
= 2*0.718 ( 750 - 300 ) = 1615.5 kJ
W3-4 = 1615.5 kJ
heat transfer
Q2-3 = W2-3 = 60KJ
Q3-4 = 0
D ) sketch of the cycle on p-V coordinates
attached below
Answer:
Density of 127 I = 
Also, 
Explanation:
Given, the radius of a nucleus is given as
.
where,
- A is the mass number of the nucleus.
The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

For the nucleus 127 I,
Mass, M = 
Mass number, A = 127.
Therefore, the density of the 127 I nucleus is given by

On comparing with the density of the solid iodine,

You can't see beyond a blind turn, so a mirror would allow you to see around the corner.
Answer:
Ff = 19.6 N
Explanation:
So since its saying whats the minimum F to move the block, we will use static friction (0.5).
We will use the equation for force of friction, which is Ff = uFn
Ff = (0.5)(4)(9.8)
Ff = 19.6 N
this is the minumum force needed to move the block, as that is the frictional force. You would need to apply a minimum force of 19.6 N to move the block