Answer:
Explanation:
As we know that loop is placed in YZ plane and magnetic field is along x direction
So here net force on the side of the loop which lies along Y axis is given as
here we know that on Y axis z = 0
so B = 0
so we have
now on the opposite side we have z = a
so magnetic field is given as
so force on that side is given as
so net force on the loop is given as
You have to draw a mathematical spatial axes .in order to judge is it right or not .. well you have to draw the crest and trough both of 1 cm in length and the total wavelength (same phase on the wave of 2 cm ) something like this
1. Domain, kingdom, phylum, class, order, family, genus, species, and subspecies.
2. If a variety of the species with a certain gene is vulnerable to a specific disease, predator, or environmental development, other members of the species with variants of the gene may be better equipped to handle the problem, and the species won't die out, just a portion of the disadvantaged.
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N