Percentage Deviation (using Eq. 8) Time of fall Deviation Trial t(s) d; = t; - tave ld P.D., = 4 x 100 P.D.1 P D., = 0.709% -0.003| 0.423 x 100 1 di = 0.420 – 0.423 =- 0.003 0.420 d2 = t2 - tave d, =...


Please check if my answers are correct. The given is the first two columns while my answers are in the third and fourth column


Percentage Deviation<br>(using Eq. 8)<br>Time of fall<br>Deviation<br>Trial<br>t(s)<br>d; = t; - tave<br>ld<br>P.D., = 4 x 100<br>P.D.1<br>P D., = 0.709%<br>-0.003|<br>0.423<br>x 100<br>1<br>di = 0.420 – 0.423<br>=- 0.003<br>0.420<br>d2 = t2 - tave<br>d, = 0.409 – 0.423<br>d, =- 0.014<br>P D. = x 100<br>PD =<br>P.D., = 3.31%<br>P.D.3=<br>2<br>I-0.014<br>0.423<br>х 100<br>0.409<br>x 100<br>dz = t3 - tave<br>dz = 0.419 – 0.423<br>dz =- 0.004<br>3<br>I-0.004<br>0.423<br>х 100<br>P.D.3<br>|P.D = 0.946%<br>0.419<br>P.D. =<br>10.007<br>P.D.4<br>х 100<br>4<br>d, = 0.430 – 0.423<br>d4 = 0.007<br>sAD - 1 = *p<br>х 100<br>%3D<br>0.423<br>0.430<br>P.D.4 = 1.65%<br>P.D.3 =<br>х 100<br>ds = ts - tave<br>ds = 0.419 – 0.423<br>d4 =- 0.004<br>5<br>P.D, =<br>-0.004<br>0.423<br>x 100<br>0.419<br>P.D. = 0.946%<br>

Extracted text: Percentage Deviation (using Eq. 8) Time of fall Deviation Trial t(s) d; = t; - tave ld P.D., = 4 x 100 P.D.1 P D., = 0.709% -0.003| 0.423 x 100 1 di = 0.420 – 0.423 =- 0.003 0.420 d2 = t2 - tave d, = 0.409 – 0.423 d, =- 0.014 P D. = x 100 PD = P.D., = 3.31% P.D.3= 2 I-0.014 0.423 х 100 0.409 x 100 dz = t3 - tave dz = 0.419 – 0.423 dz =- 0.004 3 I-0.004 0.423 х 100 P.D.3 |P.D = 0.946% 0.419 P.D. = 10.007 P.D.4 х 100 4 d, = 0.430 – 0.423 d4 = 0.007 sAD - 1 = *p х 100 %3D 0.423 0.430 P.D.4 = 1.65% P.D.3 = х 100 ds = ts - tave ds = 0.419 – 0.423 d4 =- 0.004 5 P.D, = -0.004 0.423 x 100 0.419 P.D. = 0.946%

Jun 08, 2022
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