大学物理实验绪论答案

实验绪论课作业答案

1、(1) 44.02 m (2) 12.40±0.01 m (3) 13.59 kg

(4) 0.289m (5) 5.27 m (6) 5.61×104 m 2

3、V = 4

πh •d 2 = 1.963×10-4 m 3

方法一 ：∆V = |h V ∂∂|∆h +|d V ∂∂|∆d = 4π(d 2 •∆h + 2h •d •∆d) =0.01×10-4 m 3

方法二 ：V V ∆ = h h ∆ + d

d ∆2 = 0.005 ∆V = 0.005 • V= 0.01×10-4 m 3

V = ( 1.96 ± 0.01 )×10-4 m 3 B =

V

V ∆= 0.5% P=99.7% 方法三 ：方和根传递法

∆V=()()2222d d V h h V ∆⎪⎭⎫ ⎝⎛∂∂+∆⎪⎭⎫ ⎝⎛∂∂ =4π()()()22242d hd h d ∆+∆=0.008×10-4 m 3

V = ( 1.963 ± 0.008) ×10-4 m 3 B =

V V ∆= 0.4% P=99.7% 4、 (1) ±0.5 mm (2) ±1℃ (3) ±0.01g

(4) ±0.05 mm (6) ±1.5 mA (或 ±2 mA)

5、 L =1.449×10-2 m

L σ=())110(102--∑i i L L = 0.004×10-2 m

L=(1.449±0.004) ×10-2 m B = 0.3% P =68.3% L=(1.45±0.01) ×10-2 m B = 0.7% P =99.7%

6、 R x =2

1R R •R O = 92.5Ω

①. 方法一 ：∆R x = |

1R R x ∂∂|∆R 1 +|2R R x ∂∂|∆R 2 +|0R R x ∂∂|∆R 0 = 2R R o ∆R 1 +22

1R R R o

∆R 2+21R R ∆R 0 =0.0925+ 0.0925+ 0.1 = 0.3Ω

方法二 ：x x R R ∆ = 11R R ∆ + 22R R ∆+ 00R R ∆ = 0.003 ∆R x = 0.003•R x = 0.3Ω

R x = 92.5±0.3Ω B = 0.3% P = 99.7%

②. ∆R x =()()()202022222121R R R R R R R R R x x x

∆⎪⎪⎭⎫ ⎝⎛∂∂+∆⎪⎪⎭⎫ ⎝⎛∂∂+∆⎪⎪⎭⎫ ⎝⎛∂∂=0.16Ω ≈0.2Ω

R x = 92.5±0.2Ω B = 0.2% P = 99.7%

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