The tension in the cable is mathematically given as
T =266.659 N
<h3>What is the tension in the cable.?</h3>Generally, the equation for the angle of the boom with horizontal is mathematically given as
A = tan-1(5 /10)
The angle of cable with horizontal
B = tan-1(4 / 10)
B= 21.80 degrees
Hence
175.5 * cos26.57 + 94.7 *cos26.57 * 1/2 = T (sin(26.57+21.80)) * 1
T =266.659 N
In conclusion, the tension in the cable.
T =266.659 N
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Answer:
200 mL
Explanation:
Given that,
Initial volume, V₁ = 300 mL
Initial pressure, P₁ = 0.5 kPa
Final pressure, P₂ = 0.75 kPa
We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :
V₂ is the final volume
So, the final volume of the sample is 200 mL.