Answer: 110000
Explanation:
26/9=30.5555555556
30.5555555556 x 60=1833.33333333
110000 x 60=110000
Nulceus - recipe book/instruction manual
Mitochondria - Battery
Endoplasmic reticulum - Printer or a pen?
Golgi aparatus - an envelope
Chloroplasts - green rechargable battery
Cell membrane (elastic band (2 to represent the phospholipid bilayer)
Ribosomes - I guess maybe an ink pot as its the material thats used to write
Cell Wall - the paper bag
lysosomes - washing up liquid (breaks down wate food on a dirty plate)
vaculoes - bottle of water
Acceleration = v-u/t
= (45-30)/15
= 1 km/h
Distance = ut + 1/2 at^2
s = [30 x 15] + 1/2 x 1 x 15^2
= 562.5 km
Hope this helps
Answer:
The temperature of the steam during the heat rejection process is 42.5°C
Explanation:
Given the data in the question;
the maximum temperature T
in the cycle is twice the minimum absolute temperature T
in the cycle
T
= 0.5T
now, we find the efficiency of the Carnot cycle engine
η
= 1 - T
/T
η
= 1 - T
/0.5T
η
= 0.5
the efficiency of the Carnot heat engine can be expressed as;
η
= 1 - W
/Q
where W
is net work done, Q
is is the heat supplied
we substitute
0.5 = 60 / Q
Q
= 60 / 0.5
Q
= 120 kJ
Now, we apply the first law of thermodynamics to the system
W
= Q
- Q
60 = 120 - Q
Q
= 60 kJ
now, the amount of heat rejection per kg of steam is;
q
= Q
/m
we substitute
q
= 60/0.025
q
= 2400 kJ/kg
which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)
q
= h
= 2400 kJ/kg
now, at h
= 2400 kJ/kg from saturated water tables;
T
= 40 + ( 45 - 40 ) (
)
T
= 40 + (5) × (0.5)
T
= 40 + 2.5
T
= 42.5°C
Therefore, The temperature of the steam during the heat rejection process is 42.5°C