Answer:
1.28 x 10^4 N
Explanation:
m = 1500 kg, h = 450 km, radius of earth, R = 6400 km
Let the acceleration due to gravity at this height is g'
g' / g = {R / (R + h)}^2
g' / g = {6400 / (6850)}^2
g' = 8.55 m/s^2
The force between the spacecraft and teh earth is teh weight of teh spacecraft
W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N
Answer: The temperature of the water falls by 3.3°C
Explanation:
The heat change is related to the change in temperature by the equation
dH = m Cp dT
In this example, -2665 J = 193 g x 4.184 J/g°C x dT
so dT = -3.3 °C
Answer:
can you put on a clearer image this one is hard to see
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