Which of the following is most characteristic of OSHA's relationship with professional trade organizations in residential constr
1 answer:
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Answer:
0.981
Explanation:
velocity of cars ( v1 , v2 ) = 55 mi/h
coefficient of adhesion ( u ) = 0.75
Reaction time of driver of car 1 = 2.3 -s
Reaction time of driver of car 2 = 1.9 -s
breaking efficiency of car 2 ( n2 ) = 0.80
<u>Determine the braking efficiency of car 1</u>
First determine the distance travelled during reaction time ( dr )
dr = v * tr ------- ( 1 )
tr ( reaction time )
v = velocity
note : 1 mile = 1609 m , I hour = 60 * 60 secs
<em>back to equation 1</em>
for car 1
dr1 =( 55 * 2.3 * 1609 ) / ( 60 * 60 )
= 56.53 m
for car 2
dr2 = ( 55 * 1.9 * 1609 ) / ( 60 * 60 )
= 46.70 m
<em>next we calculate the stopping distances ( d ) using the relation below</em>
ds = d + dr
d = distance travelled during break
dr = distance travelled during reaction time
where : d =
<em>for car 1 </em>
d1 =
∴ d1 =
<em>for car 2 </em>
d2 =
∴ d2 = 51.38
since the stopping distance for both cars are the same
d1 + dr1 = d2 + dr2
( 41.10 /n1 ) + 56.53 = 51.38 + 46.70
solve for n1
hence n1 = 0.981 ( braking efficiency of car 1 )
Answer:
A) ΔS_refrigerant = 0.70754 Kj/K
B) ΔS_space = -0.68441 Kj/K
C) ΔS_total = 0.02313 Kj/K
Explanation:
A) From he table attached, at Pressure of 140 KPa, and by interpolation, we get, Temperature of T = -18.77°C
Converting to degree kelvin yields;
T = -18.77 + 273 = 255.23 K
Formula for entropy change of refrigerant is given as;
ΔS_refrigerant = Q_in/T_refrigerant
We are given Q = 180 KJ
Thus, ΔS_refrigerant = 180/255.23 = 0.70754 Kj/K
B) Formula for entropy change of cooled space is given as;
ΔS_space = Q_out/T_s pace
T_space = -10°C = 273 - 10 = 263K
Thus, ΔS_space = -180/263 = -0.68441 Kj/K
C) the total entropy change would be;
ΔS_total = ΔS_refrigerant + ΔS_space
Thus,
ΔS_total = 0.70754 - 0.68441 = 0.02313 Kj/K
