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一只单摆,在第一个星球表面上的振动周期为T1;在第二个星球表面上的振动周期为T2。若这两个星球的质量之比M1∶M2=4∶1,半径之比R1∶R2=2∶1,则T1∶T2等于()
- A、1:1
- B、2:1
- C、4:1
- D、2√2:1
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更多 “一只单摆,在第一个星球表面上的振动周期为T1;在第二个星球表面上的振动周期为T2。若这两个星球的质量之比M1∶M2=4∶1,半径之比R1∶R2=2∶1,则T1∶T2等于()A、1:1B、2:1C、4:1D、2√2:1” 相关考题
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