Answer:
Neither A or B
Explanation:
The 37.3mv is not the signal voltage
sensor ground circuit does not has excessive resistance.
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
It would be 78 becuase the 2 percent of 64 is many of energy lost
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
For the first one, leaves is matter.
for the second one, ability to react
hope this helps