When both particles, the electron and the proton move at the same speed, they may have differences with their de Broglie wavelength, the particle that would have a longer wavelength would be the proton since the wavelength is in direct proportionality with the mass of the particle.
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
<h2>distance = 523 cm</h2>
Explanation:
( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s
= 5/12 rev/sec
( b ) The definition of frequency is the number of rotations per second .
Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz
( c ) The tangential speed is v = angular velocity x radius of rotation
The angular velocity ω = 2π x n , where n is the number of rotations per second
Thus angular velocity = 2π x 5/12 = 5π/6 rad/sec
The linear velocity = angular velocity x distance from center of record
Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec
Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad
Linear displacement = angular displacement x distance from center of record
= 50π/3 x 10 = 500π/3 = 523 cm
Answer:
Maybe put them in order ????
Explanation: