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旋風仔仔 2016-12-4 12:33 AM

Science 問題

Q1 A 0.3kg tennis ball travelling at 12m/s is hit by a racket and returns at 18m/s.The ball is in contact with the racket for a time of 0.1s.
(a)find the change in momentum of the ball.
(0.3*18)-(0.3*12)=1.8 kg m/s
(b)find the average force exerted in the ball by the racket.點計求教
(c)find the average force exerted in the ball on the racket.
F=(0.3*18)-(0.3*12)/0.1=18N
Q7 A vehicle mass of mass 800kg stands on an incline of 10度.If the hand brake is released,find the velocity of the vehicle after travelling 50 m down the inclined where the friction is 30N.求指教
mv=mgh(sin10)+30
800v=[800*10*50*(sin 10)]+30
v=86.83 m/s

hkpal 2016-12-4 12:26 PM

[quote]原帖由 [i]旋風仔仔[/i] 於 2016-12-4 12:33 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=452622788&ptid=26286925][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
Q1 A 0.3kg tennis ball travelling at 12m/s is hit by a racket and returns at 18m/s.The ball is in contact with the racket for a time of 0.1s.
(a)find the change in momentum of the ball.
(0.3*18)-(0.3*12)=1.8 kg m/s[/quote]
A 0.3kg tennis ball travelling at 12m/s is hit by a racket  --> Negative direction at velocity -12m/s
and returns at 18m/s --> Positive direction at velocity +18m/s

[quote] (b)find the average force exerted in the ball by the racket.點計求教[/quote]
Force x Contact time = Change in momentum

[quote] (c)find the average force exerted in the ball on the racket.
F=(0.3*18)-(0.3*12)/0.1=18N[/quote]
Action equals reaction but in opposite direction.

[quote] Q7 A vehicle mass of mass 800kg stands on an incline of 10度.If the hand brake is released,find the velocity of the vehicle after travelling 50 m down the inclined where the friction is 30N.求指教
mv=mgh(sin10)+30
800v=[800*10*50*(sin 10)]+30
v=86.83 m/s [/quote]
First calculate the net force acting on the vehicle along the inclined plane.  This is:
mg(sin10) [downplane] - 30 [upplane] = ma, where m=800kg and a = net acceleration.

Then use v2 = u2 + 2as where s = 50m and u=0.
[Alternatively, one can use work done (ma)s = change in kinetic energy 0.5mv2 - 0.5mu2]

[[i] 本帖最後由 hkpal 於 2016-12-4 12:28 PM 編輯 [/i]]
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