a weight of 1 kg is suspended from the lower end of a wire of crosssection 10mm^2 find the magnitude and direction of stress produced on it ( g=9.8 ms^_1 ) give relevant and correct answers otherwise your answers would be reported
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Step-by-step explanation:
Given -
m = 1kg
A = 10 mm2 = 10^-5 m2
g = 9.8 m/s2
F = ?
T = ?
Solution -
For suspended weight,
F = mg
F = 1×9.8
F = 9.8 N
Stress produced is given by,
T = F/A
T = 9.8/(10⁻⁵
T = 9.8 × 10⁵ Pa
T = 980 kPa
Stress produced is 980 kPa in vertically upward direction.
Answered by
1
Answer:
considering....
given,
m=1kg
A=10mm
2=10^-5
g=9.8m/s 2
F=?
T=?
solution-for suspended weight
F=mg
F=1×9.8
F=9.8n
stress produced is given by,
T=F/A
T=9.8/(10^-5)
T=9.8×10^5pa
T=980kpa
stress produced is 980 kpa in vertically upward ⬆️ direction
hope it will help you and you can understand it
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