6 000767 135×1027 8 944×10-31 9 π3．判断正误并改正错误之处。1 620010±001410 2 1254±043 78±0486 4 05m＝500mm5 31704±201 6 25355±0027 8931±0107 8 L＝20500±4×102km4．用感量e  20mg 物理天平测量某物体的质量其数据为：6135、6137、6132、6140、6133

(6) 0.0076 - Correct (7) 1.35×10^27 - Correct (8) 9.44×10^-31 - Correct (9) π - Correct

(1) 62.0010±0.01410 - Correct (2) 1.254±0.4 - Incorrect, should be 1.254±0.004 (3) 7.8±0.486 - Incorrect, should be 7.8±0.05 (4) 0.5m＝500mm - Correct (5) 31704±201 - Correct (6) 25.355±0.02 - Incorrect, should be 25.355±0.002 (7) 8.931±0.107 - Correct (8) L＝(20500±4×10^2)km - Incorrect, should be L＝(20500±400)km

4. The mass of the object can be calculated by taking the average of the measurements: (61.35+61.37+61.32+61.40+61.33+61.38)/6 = 61.3583 g. The result expression is: m = 61.3583 ± 0.01 g.

5. The density of the cylinder can be calculated using the formula: density = mass/volume. The volume can be calculated using the formula: volume = π*(d/2)^2h. Plugging in the values, we get: density = (123.75 ± 0.02)g / (π((20.146 ± 0.005)/2)^2*(50.12 ± 0.04))mm^3. The result expression is: density = (123.75 ± 0.03)g/mm^3.

(1) 62.2 + 0.0054 = 62.2054 (2) 273.15 - 27 = 246.15 (3) d = 1.20×10^-2m, S = 1/d^2 = 1/(1.20×10^-2)^2 = 6944.44 m^-2 (4) 30.00 + 21.01 - 11.5 = 39.51 (5) 100/(321 + 0.007) = 0.3119 (6) 3.04 - Correct (7) 3.000 - Correct (8) ln(98.00) - Correct (9) 104.80 - Correct