A force of (1000±0.5) N is applied on a surface of area (20+ 0.1)m².find pressure on surface.
Answers
Step-by-step explanation:
Representing error with ∆
Given :
Force = 1000N
∆F = 0.5N
Area = 20m²
∆A = 0.1 m²
We know that,
Pressure(P) = Force(F)/Area(A)
=> Applying log on both sides,
=> Log(P) = Log(F) - Log(A)
[Since, Log(a) - log(b) = log(a/b)]
Applying relative errors,
=> ∆P/P = ∆F/F + ∆A/A
Why (+)?, Everytime anyone asks for error in something, One must report the maximum error possible.
=> ∆P/P = (0.5/1000) + (0.1/20)
=> ∆P/P = 0.0005 + 0.005
=> ∆P/P = 0.0055
P = F/A,
=> P = 1000/20 = 50
=> ∆P = 50*0.0055 = 0.275
=> Pressure = 50±0.275
To get the minimum error, You can simply put a minus in the calculation in place of +
Therefore the pressure = 50±0.275
Any query? comment down below,.
Thanking you, Bunti 360 !
Step-by-step explanation:
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Secondary School Math 20 points
A force of (1000±0.5) N is applied on a surface of area (20+ 0.1)m².find pressure on surface.
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Answers
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Bunti360
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Step-by-step explanation:
Representing error with ∆
Given :
Force = 1000N
∆F = 0.5N
Area = 20m²
∆A = 0.1 m²
We know that,
Pressure(P) = Force(F)/Area(A)
=> Applying log on both sides,
=> Log(P) = Log(F) - Log(A)
[Since, Log(a) - log(b) = log(a/b)]
Applying relative errors,
=> ∆P/P = ∆F/F + ∆A/A
Why (+)?, Everytime anyone asks for error in something, One must report the maximum error possible.
=> ∆P/P = (0.5/1000) + (0.1/20)
=> ∆P/P = 0.0005 + 0.005
=> ∆P/P = 0.0055
P = F/A,
=> P = 1000/20 = 50
=> ∆P = 50*0.0055 = 0.275
=> Pressure = 50±0.275
To get the minimum error, You can simply put a minus in the calculation in place of +
Therefore the pressure = 50±0.275
Any query? comment down below,.
Thanking you, Bunti 360 !