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1. Help pls question 2b part 2

The-Spartan
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2. AQA-PHYA2-W-QP-JAN09.PDF (335.6 KB, 196 views)
3. you know the velocity is , obviously this means it travels 8.3 metres in a second.
Now you have a distance and a time, along with a force (that you calculated in part 1, use the horizontal force)
So you can now use , energy is force times distance.
So in 1 second, the ship travels 8.3 metres, so
You then have an energy . you know power
Now you have used a time of 1 second (to calculate the distance of 8.3 metres) So basically for that case,
So basically, you could say that but im not sure if that is given to you
4. (Original post by thefatone)
Help pls question 2b part 2

The-Spartan
Power = force * velocity

so the horizontal component * velocity should be correct

Units are js-1 or w
5. (Original post by The-Spartan)
you know the velocity is , obviously this means it travels 8.3 metres in a second.
Now you have a distance and a time, along with a force (that you calculated in part 1, use the horizontal force)
So you can now use , energy is force times distance.
So in 1 second, the ship travels 8.3 metres, so
You then have an energy . you know power
Now you have used a time of 1 second (to calculate the distance of 8.3 metres) So basically for that case,
So basically, you could say that but im not sure if that is given to you
thanks so much.. i couldn't remember p=fv off the top of my head and yes it is given in the data sheet

(Original post by hellomynameisr)
Power = force * velocity

so the horizontal component * velocity should be correct

Units are js-1 or w
thanks
6. (Original post by thefatone)
thanks so much.. i couldn't remember p=fv off the top of my head and yes it is given in the data sheet
no worries . That makes it easy to remember now then
7. (Original post by The-Spartan)
no worries . That makes it easy to remember now then
stress is right?

oops i meant force over area
8. (Original post by thefatone)
stress is right?

oops i meant force over area
Yup force over area.
is strain.
9. (Original post by The-Spartan)
Yup force over area.
is strain.
i'm stuck on Q 4 a now, which way does velocity go???
10. (Original post by thefatone)
i'm stuck on Q 4 a now, which way does velocity go???
Velocity here is a tangent to the path at P, acting in the bottom right way.
At the start it is completely horizontal, eventually acceleration downwards will make it nearly vertical. here its in between
11. (Original post by The-Spartan)
Velocity here is a tangent to the path at P, acting in the bottom right way.
At the start it is completely horizontal, eventually acceleration downwards will make it nearly vertical. here its in between
huh??? i don't understand so is it acting alone the black line ???
12. (Original post by thefatone)
huh??? i don't understand so is it acting alone the black line ???
Yes it is. Looks sort of like this (ish)
Spoiler:
Show
13. (Original post by The-Spartan)
Yes it is. Looks sort of like this (ish)
Spoiler:
Show
yea that's what i meant

also the next question i can do question 4 b part 1
however when using SUVAT shouldn't the distance be the distance travelled not distance interms of height?
14. (Original post by thefatone)
yea that's what i meant

also the next question i can do question 4 b part 1
however when using SUVAT shouldn't the distance be the distance travelled not distance interms of height?
This is a great example of the hidden part of parabolic motion.
Time spent in the horizontal is equal to time in the vertical,
So setting , you get from like you know

Basically SUVAT works in both the horizontal and vertical components the same. Here its the vertical.

EDIT: here is a video showing this time effect
15. (Original post by The-Spartan)
This is a great example of the hidden part of parabolic motion.
Time spent in the horizontal is equal to time in the vertical,
So setting , you get from like you know

Basically SUVAT works in both the horizontal and vertical components the same. Here its the vertical.

EDIT: here is a video showing this time effect
but how??? the distance the ball hits the ground is 27m away from the tower thus the time cannot be same
16. (Original post by thefatone)
but how??? the distance the ball hits the ground is 27m away from the tower thus the time cannot be same
It can, because if you think about it, when i throw the ball the timer starts for both components. When it hits the ground it stops for both components, therefore the time is the same in horizontal and vertical.

It has two separate velocities for horizontal and vertical . This also applies for acceleration and distance travelled

The SUVATs work in both horizontal and vertical components using their respective parameters.

for horizontal and

In this case, we use vertical so , and .

Hard concept to explain, hope i helped at least abit xD
17. (Original post by thefatone)
but how??? the distance the ball hits the ground is 27m away from the tower thus the time cannot be same
Just shut up if u don't understand stuff as easy as this you're in the wrong room
18. (Original post by The-Spartan)
It can, because if you think about it, when i throw the ball the timer starts for both components. When it hits the ground it stops for both components, therefore the time is the same in horizontal and vertical.

It has two separate velocities for horizontal and vertical . This also applies for acceleration and distance travelled

The SUVATs work in both horizontal and vertical components using their respective parameters.

for horizontal and

In this case, we use vertical so , and .

Hard concept to explain, hope i helped at least abit xD
i still can't understand why the time is the same, because the ball has that little extra distance to travel going vertically instead of just horizontally thus the time take should be bigger but a really small increase compared to time taken for just dropping the ball.

For example like a segment of the circle
if you draw a straight line across to form a triangle and you have that midgy extra bit of area between the arc and straight line that to me would be the minuscule difference.

So do you see why i say the time must be different?

(Original post by STRANGER2)
Just shut up if u don't understand stuff as easy as this you're in the wrong room
If people were quiet and never asked questions the population as a whole would be much stupider because if someone was unsure of how to carry out a task their boss set them and they did it wrong things wouldn't have a good outcome.

Look i'm trying my best really to learn this stuff and understand WHY. I CAN DO IT BUT JUST BECAUSE I CAN DO IT DOESN'T MEAN THAT I UNDERSTAND IT.

The internet is not a room, however i am in the student room

19. (Original post by thefatone)
i still can't understand why the time is the same, because the ball has that little extra distance to travel going vertically instead of just horizontally thus the time take should be bigger but a really small increase compared to time taken for just dropping the ball.

For example like a segment of the circle
if you draw a straight line across to form a triangle and you have that midgy extra bit of area between the arc and straight line that to me would be the minuscule difference.

So do you see why i say the time must be different?
I completely understand what you're saying. Let me explain it in a different way...

Suppose i just dropped that ball right? The acceleration is obviously g downwards. So the ball falls towards the ground, accelerating at 9.81 (metres per second) per second. The only acceleration acting downwards is this, and therefore you can use SUVATs to find time.

Now lets consider that i throw the ball, from the same height. The acceleration acting down on the ball is still only g. This means that the ball, as before, is accelerating towards the ground at 9.81 (metres per second) per second. Therefore the only acceleration acting downwards is this, like the dropping scenario, as they are both dropped from the same height they both hit the ground at the same time

Now i hear you saying in your head, oh but does the one you throw not travel further and so must be in the air for longer?

No The balls both hit the ground when their vertical distance travelled is 24m right?
If they are both accelerating towards the ground at 9.81ms^-2, then they both reach the ground at the same time. The horizontal velocity of the ball is not a factor. Im going to use some diagrams to show this effect visually, where i show the distance travelled in a time.

This is me dropping the ball from a height right? now lets compare to if i threw that ball:

Can you see how they travel the same vertical distance in the same time?
This shows that the time spent in the air is the same in both cases from the same height. Hope this helped.

Edit:
(Original post by STRANGER2)
Just shut up if u don't understand stuff as easy as this you're in the wrong room
You're not helping. This stuff is not 'easy'. Anyway this is the maths science and technology academic help section.
If you're not going to help, GTFO.
20. (Original post by The-Spartan)
I completely understand what you're saying. Let me explain it in a different way...

Suppose i just dropped that ball right? The acceleration is obviously g downwards. So the ball falls towards the ground, accelerating at 9.81 (metres per second) per second. The only acceleration acting downwards is this, and therefore you can use SUVATs to find time.

Now lets consider that i throw the ball, from the same height. The acceleration acting down on the ball is still only g. This means that the ball, as before, is accelerating towards the ground at 9.81 (metres per second) per second. Therefore the only acceleration acting downwards is this, like the dropping scenario, as they are both dropped from the same height they both hit the ground at the same time

Now i hear you saying in your head, oh but does the one you throw not travel further and so must be in the air for longer?

No The balls both hit the ground when their vertical distance travelled is 24m right?
If they are both accelerating towards the ground at 9.81ms^-2, then they both reach the ground at the same time. The horizontal velocity of the ball is not a factor. Im going to use some diagrams to show this effect visually, where i show the distance travelled in a time.

This is me dropping the ball from a height right? now lets compare to if i threw that ball:

Can you see how they travel the same vertical distance in the same time?
This shows that the time spent in the air is the same in both cases from the same height. Hope this helped.

Edit:

You're not helping. This stuff is not 'easy'. Anyway this is the maths science and technology academic help section.
If you're not going to help, GTFO.
^^ you've perfectly demonstrated why i think the path where the ball is thrown is longer since you've just made a triangle, the path where the ball is thrown looks like the longest side, if the ball has more distance to travel the time taken cannot be the same.

But why is horizontal velocity not counted? that's because we're doing suvat for vertical you can use for horizontal

well ffs then -.-

Lol thanks for backing me up.. dunno what's with that guy :/
21. (Original post by The-Spartan)
x.
Help with 5 a pls i have no idea about waves >.> they're my worst

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