What type of material is wrapped around an object could cause the object to lose the most heat
1 answer:
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Answer:
None of the above forces on air drag on him is equal to his weight
Explanation:
In the velocity-time graph,the gradient of the curve where it is flatten shows the parachutist reaches the terminal velocity when it reaches terminal velocity which means the parachutist reaches constant velocity or speed,indicating that the acceleration of free fall(g) is zero.And according to the resultant force formula weight - air drag= mass*acceleration. so when accelerate is zero,resultant force is zero. And hence the equation will be like this: weight= air drag
Answer:
1 m
Explanation:
L = 100 m
A = 1 mm^2 = 1 x 10^-6 m^2
Y = 1 x 10^11 N/m^2
F = 1000 N
Let the cable stretch be ΔL.
By the formula of Young's modulus
ΔL = 1 m
Thus, the cable stretches by 1 m.
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
