题目内容
(请给出正确答案)
[不定向]
某钢尺的尺长方程式为:L=30-0.003+1.25×10-5×30×(t-20℃),若用其在4℃的温度下丈量了60m,则其温度改正数和实际长度分别为()。
A、-0.012m;59.988m
B、+0.012m;60.012m
C、-0.012m;59.982m
D、+0.012m;59.982m
暂无答案
A、-0.012m;59.988m
B、+0.012m;60.012m
C、-0.012m;59.982m
D、+0.012m;59.982m
A、-0.121
B、0.0424
C、0.0121
D、0
A、=30m-0.003m+1.25×10-5×(t一20°C)×30m
B、=30m+0.003m+1.25×10-5×(t一20℃)×30m
C、=30m+0.006m+1.25×10-5×(t一20℃)×30m
D、=30m-0.006m+1.25×10-5×(t一20℃)×30m
A、-0.0172
B、0.0172
C、-0.0118
D、0.0118
A、-10mm
B、+10mm
C、+20mm
D、-20mm
A、-10mm
B、+10mm
C、+20mm
D、-20mm