Answer:
a) W = 25.5 lbf
b) W = 150 lbf
Explanation:
Given data:
Mass of astronaut = 150 lbm
local gravity = 5.48 ft/s^2
a) weight on spring scale
it can be calculated by measuring force against local gravitational force which is equal to weight of body
W = mg
b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity
thus W= 150 lbf
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) 358.8K
b) 181.1 kJ/kg.K
c) 0.0068 kJ/kg.K
Explanation:
Given:
P1 = 100kPa
P2= 800kPa
T1 = 22°C = 22+273 = 295K
q_out = 120 kJ/kg
∆S_air = 0.40 kJ/kg.k
T2 =??
a) Using the formula for change in entropy of air, we have:
∆S_air =
Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K
Solving, we have:
[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]
Solving for T2 we have:
Taking the exponential on the equation (both sides), we have:
[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]
b) Work input to compressor:
= 184.1 kJ/kg
c) Entropy genered during this process, we use the expression;
Egen = ∆Eair + ∆Es
Where; Egen = generated entropy
∆Eair = Entropy change of air in compressor
∆Es = Entropy change in surrounding.
We need to first find ∆Es, since it is unknown.
Therefore ∆Es =
∆Es = 0.4068kJ/kg.k
Hence, entropy generated, Egen will be calculated as:
= -0.40 kJ/kg.K + 0.40608kJ/kg.K
= 0.0068kJ/kg.k