Answer:
Normal force = 0.326N
Explanation:
Given that:
mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg
height of the mass = 1.1 m
radius = 0.2 m
acceleration due to gravity = 9.8 m/s²
We are to determine the normal force pressing on the track at A.
To to that;
Let consider the conservation of energy relation; which says:
mgh = mgr + 1/2 mv²
gh = gr + 1/2 v²
gh - gr = 1/2v²
g(h-r) = 1/2v²
v² = 2g(h-r)
However; the normal force will result to a centripetal force; as such, using the relation
N =mv²/r
replacing the value for v² = 2g(h-r) in the above relation; we have:
Normal force = 2mg(h-r)/r
Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2
Normal force = 0.065268/0.2
Normal force = 0.32634 N
Normal force = 0.326N
Where is the following steps??
The rotor is situated inside the stator and is mounted on the AC motor's shaft. It is the rotating part of the AC motor. And while we know this, the major function of the rotor and the stator is helping the motor shaft rotate.
Answer:
umax = 0.1259ft/s
Explanation:
Given:
•Distance between plates, B = 0.01ft
•Pressure difference decrease, 
•Fluid viscosity, u = 10^-³lbf-s/ft²
Specific gravity, S = 0.80
Max velocity in the z-direction will be:
![u_max= [\frac{B^2y}{8u}]\frac{dh}{ds}](https://tex.z-dn.net/?f=u_max%3D%20%5B%5Cfrac%7BB%5E2y%7D%7B8u%7D%5D%5Cfrac%7Bdh%7D%7Bds%7D)
Substituting for h in the first equation, we have:
![\frac{d}{dz}[\frac{p}{y}+z]](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdz%7D%5B%5Cfrac%7Bp%7D%7By%7D%2Bz%5D)
= -0.20192
Substituting dh/dz value in the first equation (umax), we have:
umax = 0.1259ft/s
Answer:
79 kW.
Explanation:
The equation for enthalpy is:
H2 = H1 + Q - L
Enthalpy is defined as:
H = G*(Cv*T + p*v)
This is specific volume.
The gas state equation is:
p*v = R*T (with specific volume)
The specific gas constant for air is:
287 K/(kg*K)
Then:
T1 = 60 + 273 = 333 K
T2 = 200 + 273 = 473 K
p1*v1 = 287 * 333 = 95.6 kJ/kg
p2*v2 = 287 * 473 = 135.7 kJ/kg
The Cv for air is:
Cv = 720 J/(kg*K)
So the enthalpies are:
H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW
H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW
Ang the heat is:
Q = 34 kW
Then:
H2 = H1 + Q - L
381 = 268 + 34 - L
L = 268 + 34 - 381 = -79 kW
This is the work from the point of view of the air, that's why it is negative.
From the point of view of the machine it is positive.
