Data Sheet B v\' vs. l/m (r,R are constant) constant value of r = 049S ± 0002 m

Question

Data Sheet B v' vs. l/m (r,R are constant) constant value of r = 049S ± 0002 m constant value of Fe=-Szq_ ± ,5g8 N starting value of m =_OSS kg… 0. 0792 Expected Trend for T as Mass Increases: T should (increase. decrease) (circle one) 0-403 d //0002 t 0.0. ) Data Table B total mass m (mean value) of spinning (kg) (sec) (m/s) mass m deu 3.2779 3.1.34 xlo 0.403 2.48 ±0.02|1.8013 ±007:4 |1 e 2 195 ± O oss |3.246 |2630 O.800 | | ±01s1 3 kg | 2 208 | 0.7554 ±0.02 ±0.193 oss s kg |1908 |08448 ±0.02 |1KM ±00412 104 ±0.141|2.545 " (v2) = [2(8v/v)v2], which simplifies to: (v)-ry(Ev)] 2· The slope of the straight line forv2 vs.m½ .) Calculated value of Fe* below.) 3. -(-). (Show calculations 4. What is the percent error of the two values?

Explanation / Answer

form the given data

by plotting on graph v^2 vs per m

we get

a. slope = 0.5987 +- 0.77% = 0.5987 +- 0.0046099

b. Fc = 5.39 +- 0.58 8N

r = 0.195 +- 0.002 m

hence

Fc.r = 1.05105

d(Fc.r) = Fc.r(d(Fc)/Fc + dr/r) = Fc.r(0.588/5.39 + 0.002/0.195) = 0.12544 Nm

hence

Fc.r = 1.05105 +- 0.12544 Nm

now, slope = mv^2 = Fc.r

calculated Fc.r = 0.5987 +- 0.0046099

c. %error = 43.037 %

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