Answer:
Hence, the three effects of electric current are heating effect, magnetic effect and chemical effect.
Answer:
a) The mechanical force is -226.2 N
b) Using the coenergy the mechanical force is -226.2 N
Explanation:
a) Energy of the system:
![\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B1.2%2Ai%5E%7B1%2F2%7D%20%7D%7Bg%7D%20%5C%5Ci%3D%28%5Cfrac%7B%5Clambda%20g%7D%7B1.2%7D%20%29%5E%7B2%7D)
![\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cdelta%20w_%7Bf%7D%20%7D%7B%5Cdelta%20g%7D%20%3D%5Cfrac%7Bg%5E%7B2%7D%5Clambda%20%5E%7B3%7D%20%20%7D%7B3%2A1.2%5E%7B2%7D%20%7D)
![f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }](https://tex.z-dn.net/?f=f_%7Bm%7D%3D-%20%5Cfrac%7B%5Cdelta%20w_%7Bf%7D%20%7D%7B%5Cdelta%20g%7D%20%3D-%5Cfrac%7Bg%5E%7B2%7D%5Clambda%20%5E%7B3%7D%20%20%7D%7B3%2A1.2%5E%7B2%7D%20%7D)
If i = 2A and g = 10 cm
![\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B1.2%2Ai%5E%7B1%2F2%7D%20%7D%7Bg%7D%20%3D%5Cfrac%7B1.2%2A2%5E%7B1%2F2%7D%20%7D%7B10x10%5E%7B-2%7D%20%7D%20%3D16.97)
![f_{m}=-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N](https://tex.z-dn.net/?f=f_%7Bm%7D%3D-%5Cfrac%7Bg%5E%7B2%7D%5Clambda%20%5E%7B3%7D%20%20%7D%7B3%2A1.2%5E%7B2%7D%20%7D%3D-%5Cfrac%7B16.97%5E%7B3%7D%2A2%2A0.1%20%7D%7B3%2A1.2%5E%7B2%7D%20%7D%20%3D-226.2N)
b) Using the coenergy of the system:
![f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2} }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N](https://tex.z-dn.net/?f=f_%7Bm%7D%3D-%20%5Cfrac%7B%5Cdelta%20w_%7Bf%7D%20%7D%7B%5Cdelta%20g%7D%20%3D-%5Cfrac%7B1.2%2A2%2Ai%5E%7B3%2F2%7D%20%20%7D%7B3%2Ag%5E%7B2%7D%20%7D%3D-%5Cfrac%7B1.2%2A2%2A2%5E%7B3%2F2%7D%20%7D%7B3%2A0.1%5E%7B2%7D%20%7D%20%3D-226.2N)
Answer:
work done = 48.88 ×
J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 ×
J
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m