Technician A says that a fully charged battery is less likely to freeze than a discharged battery. Technician B says that the st
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Clockwise torque due to 100g is 0.1029 Nm and 200g is 1.4406 Nm. Clockwise torque due to stick mass is 0.2254 Nm and Counter-clockwise torque due to normal force is 1.7689 Nm.
<h3>What is clockwise torque?</h3>The right-hand rule for cross products determines the direction of torque, which is calculated as the cross product of force and distance. Your thumb will point in the direction of the torque if you place your palm in the direction of the applied force and extend your fingers from the pivot point in that direction.
A related right-hand rule relates the direction of the rotation to the direction of the torque. Your fingers will curl in the direction of rotation if you point your thumb in the direction of the torque.
Positive torques cause counter clockwise rotation, while negative torques cause clockwise rotation.
The sum of all torques must be zero at equilibrium since an object in equilibrium has no net torque.
When the force is applied in a direction perpendicular to the line connecting the pivot and the force, the torque is at its greatest.
You can calculate the torque's magnitude using
To solve problems involving torques, follow these eight steps: read the issue, create a free-body diagram, locate the pivot point, write down the expressions for all torques, For equilibrium conditions, set the sum of torques to zero, list all known variables, pick the desired variable(s), write down equations involving those variable(s), solve the equations, plug in numbers, and test your solution.
Clockwise torque due to 100 g ⇒ T1 = 0.105* 9.8* 0.1 = 0.1029 Nm
Clockwise torque due to 200 g ⇒ T2 = 0.210* 9.8* 0.7 = 1.4406 Nm
Clockwise torque due to stick mass ⇒ T3 = 0.046* 0.5* 9.8 =0.2254 Nm
Counter-clockwise torque due to normal force ⇒ T4 = (0.046 + 0.21 + 0.105)*9.8* 0.5 = 1.7689 Nm
Learn more about torque
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Answer:
a) 764.45K
b) 210.48 kJ/kg
c) 30.14%
Explanation:
pressure ratio = 10
minimum temperature = 295 k
maximum temperature = 1240 k
isentropic efficiency for compressor = 83%
Isentropic efficiency for turbine = 87%
<u>a) Air temperature at turbine exit </u>
we can achieve this by interpolating for enthalpy
h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air
T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) = 764.45K
<u>b) The net work output </u>
first we determine the actual work input to compressor
Wc = h2 - h1 ( calculated values )
= 626.57 - 295.17 = 331.4 kJ/kg
next determine the actual work done by Turbine
Wt = h3 - h4 ( calculated values )
= 1324.93 - 783.05 = 541.88 kJ/kg
finally determine the network output of the cycle
Wnet = Wt - Wc
= 541.88 - 331.4 = 210.48 kJ/kg
<u>c) determine thermal efficiency </u>
лth = Wnet / qin ------ ( 1 )
where ; qin = h3 - h2
<em>equation 1 becomes </em>
лth = Wnet / ( h3 - h2 )
= 210.48 / ( 1324.93 - 626.57 )
= 0.3014 = 30.14%
Answer:
The heater load =35 KJ/kg
Explanation:
Given that
At initial condition
Temperature= 15°C
RH=80%
At final condition
Temperature= 50°C
We know that in sensible heating process humidity ratio remain constant.
Now from chart
At temperature= 15°C and RH=80%
At temperature= 50°C
The heater load = 73 - 38 KJ/kg
The heater load =35 KJ/kg