Science, asked by sheelathomas304, 7 months ago

calculate the maximum possible error y=4 mgL/πd^2L

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Answered by Anonymous

Answer:

Y= πD

Y= πD 2

Y= πD 2 l

Y= πD 2 l 4MgL

Y= πD 2 l 4MgL Y

Y= πD 2 l 4MgL Y ΔY

Y= πD 2 l 4MgL Y ΔY =

Y= πD 2 l 4MgL Y ΔY = L

Y= πD 2 l 4MgL Y ΔY = L ΔL

Y= πD 2 l 4MgL Y ΔY = L ΔL +

Y= πD 2 l 4MgL Y ΔY = L ΔL + D

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D +

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2×

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 +

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y ΔY

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y ΔY %=0.03024×100

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y ΔY %=0.03024×100 =3.024%

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