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:
(a) 0.17 m
(b) 5.003 m
(c) 6.38 ×
N
(d) 7.37 ×
N
Explanation:
(a) The minimum value of
will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.
(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

<em>Hence, the maximum distance is 5.002 m</em>
(c) For minimum magnitude we use the minimum distance calculated in (a)
Minimum Distance = 0.17 m
For electrostatic force= 

×
(d) For maximum magnitude, we use the maximum distance calculated in (b)
Maximum Distance = 5.002 m
Using the formula for electrostatic force again:
F = 
F= 7.37×
N
Answer:
10×2=20
10×2÷64=??
I am not sure what you are trying to say...
Answer:
answer is 3.05v
Explanation:
hope it is helpful and briliant