physics功課
2006-10-30 10:40 pm
一車剎車掣,衝行了80m才停定....假設剎掣後,汽車受路面的摩擦力為10000N.....設汽車的質量為1000kg,求汽車剎掣前的速率為多少km/h
回答 (3)
2006-11-01 4:47 am
✔ 最佳答案
設a為汽車剎掣加速率根據牛頓第二定律
F = ma
-10000 = 1000a
a = -10ms-2 (負數為減速率)
設u為汽車剎掣前的速率.
根據 v^2 = u^2 + 2as
0^2 = u^2 + 2(-10)(80)
u = 40 ms-1 = 40 X 3600 / 1000 = 144 km/h
汽車剎掣前的速率為144 km/h.
2006-11-02 5:34 am
Braking force = 10000 N
As F = ma
10000 = 1000 x a
a = 10 m s^-1 (negative acceleration)
v² - u² = 2as
-u² = 2 x -10 x 80
u = 40 m s^-1
40 m s^-1
= (40/1000) / (1 / 3600)
= 144 km/h
Therefore the speed of the car before applying brakes is 144 km/h
As F = ma
10000 = 1000 x a
a = 10 m s^-1 (negative acceleration)
v² - u² = 2as
-u² = 2 x -10 x 80
u = 40 m s^-1
40 m s^-1
= (40/1000) / (1 / 3600)
= 144 km/h
Therefore the speed of the car before applying brakes is 144 km/h
2006-10-31 2:28 am
accelaration: F=ma
10000=1000a
a=10ms^-2
v^2=2as+u^2
0=2(-10)( 80)+u^2
u=40ms^-1
10000=1000a
a=10ms^-2
v^2=2as+u^2
0=2(-10)( 80)+u^2
u=40ms^-1
收錄日期: 2021-04-13 14:45:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061030000051KK02724