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Physics 23 Online
OpenStudy (anonymous):

will somebody help me with my homework? A body slides down in a frictionless plane during the third second after starting from rest it travels 19.4 meters. What is the angle of the plane? Thanks in advance

OpenStudy (anonymous):

x= 0.5 g t^2

OpenStudy (science0229):

|dw:1407057873069:dw|

OpenStudy (anonymous):

oh i forgot to put sin theta in my equation.

OpenStudy (science0229):

Since it took 3 seconds to travel 19.4 meters, its final velocity is 19.4/3 Its initial velocity is 0, of course, since it was at rest, first.

OpenStudy (science0229):

@Careless

OpenStudy (science0229):

Using the work-energy theorem...\[Work=(Force)(Distance)=\Delta (Kinetic Energy)=\frac{ 1 }{ 2 }m {v _{f}}^2-\frac{ 1 }{ 2 }m {v _{i}}^2\]

OpenStudy (science0229):

vf is the final velocity, which is 19.4/3 vi is the initial velocity, which is 0 Distance is 19.4

OpenStudy (science0229):

Force is as following|dw:1407058991396:dw|

OpenStudy (science0229):

I'm guessing that you're using 9.8 for the gravitational acceleration (of Earth)

OpenStudy (science0229):

since we're on the inclined plane, the force would be \[9.8m \sin \theta\]

OpenStudy (science0229):

Substitute all of that in the work-energy theorem

OpenStudy (science0229):

\[9.8m \sin \theta*19.4=\frac{ 1 }{ 2 }m (\frac{ 19.4 }{ 3 })^2\]

OpenStudy (science0229):

Solve for sin to get\[\sin \theta=\frac{ 97 }{ 882 }\]

OpenStudy (science0229):

You can solve for theta on your own.

OpenStudy (science0229):

If the gravitational acceleration is 'g', then the final equation would turn out as\[\sin \theta=\frac{ 97 }{ 90g }\]

OpenStudy (science0229):

This turned out ugly...

OpenStudy (science0229):

@Catch.me Did I do something wrong? When the answer comes out like this, I always have a feeling that I made a mistake somewhere

OpenStudy (anonymous):

I didn't follow ,but the main idea is right (don't mind about small errors you just give the keys to answer it :D)

OpenStudy (science0229):

I guess so :)

OpenStudy (anonymous):

thank you :)

OpenStudy (anonymous):

an arrow is shot vertically upward with a speed of 288 fps and 3 seconds later another arrow is shot up with a speed of 244 fps. will they meet? if so, where?

OpenStudy (anonymous):

please answer this another question :)

OpenStudy (anonymous):

Hey show your work and why can't solve it?

OpenStudy (anonymous):

after i have finish my homework i will show my solution ok? I'm sorry because i am very busy right now. We have midterm this week. so stay still. until i show my solution. Thank you for understanding.

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