Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N
Answer:
The magnitudes of the net magnetic fields at points A and B is 2.66 x T
Explanation:
Given information :
The current of each wires, I = 4.7 A
dH = 0.19 m
dV = 0.41 m
The magnetic of straight-current wire :
B= μI/2πr
where
B = magnetic field (T)
μ = 1.26 x
(N/
)
I = Current (A)
r = radius (m)
the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,
BH = μI/2πr
= (1.26 x )(4.7)/(2π)(0.19)
= 4.96 x T
BV = μI/2πr
= (1.26 x )(4.7)/(2π)(0.41)
= 2.3 x T
hence,
the net magnetic field = BH - BV
= 4.96 x - 2.3 x
= 2.66 x T
Answer:
T = 60 s
Explanation:
There are 6 poles on the track which are equally spaced
so the angular separation between the poles is given as
so the angular speed of the train is given as
now we have time period of the train given as