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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
The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

Here,
= Permeability at free space
N = Number of loops
A = Cross-sectional Area
l = Length
Replacing with our values we have,



Therefore the Inductance is 
Answer:
You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over is explained below in details.
Explanation:
spinning ball halts after traveling some range due to friction energy act different direction of movement of the ball. you can observe in the figure.
Let any rolling ball of mass (m ) is traveling with velocity v ,
common effect on ball (N) = mg
because of motion, friction energy develops on the contact exterior and begins to resist the movement of the rolling ball.
hence,
fr = uN = umg act on communicating exterior, so, after any time due to friction energy rolling ball gets to rest.