The tension in the cable is 23.2 N
<h3>What is the tension in the string?</h3>
The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.
Tcosθ, is acting perpendicularly, Tcosθ = 0
Taking moments about the pivot:
Tsinθ * 2.2 = 4 * 9.8 * 0.7
Solving for θ;
θ = tan⁻¹(1.4/2.2) = 32.5°
T = 27.44/(sin 32.5 * 2.2)
T = 23.2 N
In conclusion, the tension in the cable is determined by taking moments about the pivot.
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Answer:
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Answer:
The answer would be 0.04ohms.
Explanation:
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Answer:
A) ( - 200t + 40 ) volts
B) b) anticlockwise , c) anticlockwise , d) clockwise , e) clockwise
Explanation:
Given data:
magnetic flux (Φm) = 5.0t^2 − 2.0t
number of turns = 20
<u>a) determine induced emf </u>
E = - N 
= - N ( 10t - 2 ) = - 20 ( 10t - 2 )
= - 200t + 40 volts
<u>b) Determine direction of induced current </u>
i) at t = 0
E = - 0 + 40 ( anticlockwise direction )
ii) at t = 0.10
E = -20 + 40 = 20 ( anticlockwise direction )
iii) at t = 1
E = - 200 + 40 = - 160 ( clockwise direction)
iv) at t = 2
E = -400 + 40 = - 360 ( clockwise direction )
<span>F x L = W x X whereW=weight is total load = 80, L is length from fulcrum which is the unknown and what we are solving for. x= length we know. and F equals 50 force we know. So (W*X)/F=LL equals 64</span>
