Physics paper

Help pls question 2b part 2

@The-Spartan

you know the velocity is $8.3ms^{-1}$, 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 $E=Fd$, energy is force times distance.
So in 1 second, the ship travels 8.3 metres, so $E=8.3F_h$
You then have an energy $E$. you know power $P=\frac{E}{t}$
Now you have used a time of 1 second (to calculate the distance of 8.3 metres) So basically for that case, $P=E$
So basically, you could say that $P=Fv$ but im not sure if that is given to you

Power = force * velocity

so the horizontal component * velocity should be correct

Units are js-1 or w

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

stress is $\dfrac{\Delta L}{L}$ right?

oops i meant force over area

Yup force over area.
$\dfrac{\Delta L}{L}$ is strain.

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

huh??? i don't understand so is it acting alone the black line ???

Yes it is. Looks sort of like this (ish)

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 $S=24, a=g=9.81, u=0$, you get $24=\frac{1}{2}gt^2$ from $S=ut+\frac{1}{2}at^2$ 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

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 $v_h, v_v$. This also applies for acceleration $a_h, a_v$ and distance travelled $s_h, s_v$

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

$s_h=u_ht+1/2a_ht^2$ for horizontal and $s_v=u_vt+1/2a_vt^2$

In this case, we use vertical so $s_v = 24$, $a_v=9.81$ and $u_v=0$.

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?

Just shut up if u don't understand stuff as easy as this you're in the wrong room

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.

x.