F4 Physics calculating

2009-11-30 1:07 am

John* s watch runs slower than the actual time by one minute each day. John adjusted
the watch to the correct at 2:00 pm on 1 January 2009.

1. What time should be shown on John* s watch when the actual time is 2: 00 pm on 10
Jan 2009?

2. What is the percentage error of time shown on John* s watch as compared with the
correct time?

更新1:

1. ans= 1:51 pm on 1 Jan 2009 2. 6.94 x 10^ -2 %

回答 (1)

Prof. Physics

2009-11-30 4:01 am

✔ 最佳答案

1. From 1 Jan to 10 Jan, there are 10 days. So, the time delay for John's watch is 10 minutes.

So, the time on his watch = 1.51 pm on 10 Jan 2009.

2. Total minutes in 10 days = 24 X 60 X 10 = 14 400 mins

Percentage error

= 10 / 14 400 X 100%

= 6.94 X 10^-2 %

參考: Physics king

👍 0 👎 0

收錄日期: 2021-04-19 20:50:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091129000051KK01199

檢視 Wayback Machine 備份