Final Velocity of 5.0g Bullet: 117m/s

[Ayesha Shafique]

What will the final velocity of 5.0 g bullet starting from rest if a net force of 45 N is applied over a distance of 0.80 m ?2.3. i tried to solve it but i don't know of my anwer is correct
F=ma, F/m=a
45/.005 kg
a=9000m/s^-2
From this formula ( d=1/2at^2) we can say that
Time= 2d/a and then take under root
2×0.80/9000 (under root or square root)
=0.013 sec
initial velocity (u) =0
a=v-u/t
at=v-u
9000×0.013=v-0
117+0=v
answer is 117m/s^-1

 

[Alettix]

I think this looks correct, but I would recommend you to read the guidelines for students before posting again. :)
https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/

 

[CWatters]

You get 120m/s if you don't round down to 0.013 seconds.

 

[Ayesha Shafique]

You get 120m/s if you don't round down to 0.013 seconds.

I don't understand what you mean by this. Please tell what does it mean because i am only in 8 grade

 

[Alettix]

I don't understand what you mean by this. Please tell what does it mean because i am only in 8 grade

What CWatters means is that 0,013 s is not the exact value of the time. If you use the exact value, which is t = (2*0,8/9000)^(1/2), you will get a slightly different, more correct answer. I would recommend googleing "significat numbers". :)

 

[Ayesha Shafique]

What CWatters means is that 0,013 s is not the exact value of the time. If you use the exact value, which is t = (2*0,8/9000)^(1/2), you will get a slightly different, more correct answer. I would recommend googleing "significat numbers". :)

Thank you very much for your help. I also had the same answer but the problem was that i shortened 0.0133333333333 to 0.013 that's why my answer was 117 and not 120.