When a net force of 17.0 newtons is applied to a dictionary placed on a frictionless table, it accelerates by 3.75 meters/second
1 answer:
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(a) The spring constant is 500 N/m.
(b) The extension of the spring when 25 N force is applied is 0.05 m.
(c) The applied force to cause an extension of 5 mm is 2.5 N.
The given parameters:
- Applied force, F = 10 N
- Extension of the spring, x = 20 mm
The spring constant is calculated as follows;
The extension of the spring when 25 N force is applied is calculated as follows;
The applied force to cause an extension of 5 mm is calculated as follows;
Learn more about Hook's law here: brainly.com/question/12253978
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
