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1. Just started a bit of AS revision and have no text book or good notes - got stuck on a couple of past paper questions - any ideas?

Question:

A floating fire tender is testing its pumps. The hose in use is being directed virtically upwards but the water is being blown to the left by a strong sideways wind.

Considering the path of one droplet of water:

h1 - the height above sea level of the outlet of the hose = 10m
h2 - the maximum height above sea level of the droplet = 28m

Vertical velocity of the water leaving the hose = 19 ms-1

By considering the vertical motion of the droplet and ignoring air resistance,

(a) Calculate the time taken for the droplet to reach its maximum height.
(b) Calculate the time taken for the droplet to fall from maximum height to sea level.

For (a) i just used time = distance / speed -- 28/19 = 1.5 seconds - is that right?

For (b) i'm a bit stuck. I assume that acceleration due to gravity will come into it. I dfidn't think i could use 'v = u + at' as it doesn't take height into consideration. I couldn't use 'v^2 = u^2 + 2as' as i didn't know what to use for 'v'. I tried using 'd = ut + 1/2 at^2' by changing it to '(d-u) / 1/2 a = t' but that came out as 5.71 secons and i don't think that's right. What do i do?

Question 2:

"Most electrical showers draw cold water direct from the main supply and heat it as it is used - day or night.

(a) Write a word equation to describe the energy transfer that takes place in an electric shower.
(b) Rewrite the equation using appropriate formula.

Any help is much appreciated.

EDIT - had to change some of the data on question 1 as i realised i had made a mistake with resolving the vector and had the heights wrong. - sorry
2. Which exam board is this for?
3. edexcel - it's a specimen paper
4. (Original post by need_help)
Just started a bit of AS revision and have no text book or good notes - got stuck on a couple of past paper questions - any ideas?

Question:

A floating fire tender is testing its pumps. The hose in use is being directed virtically upwards but the water is being blown to the left by a strong sideways wind.

Considering the path of one droplet of water:

h1 - the height above sea level to the outlet of the hose = 20m
h2 - the maximum height above sea level of the droplet = 56m

Vertical velocity of the water leaving the hose = 19.8 ms-1

By considering the vertical motion of the droplet and ignoring air resistance,

(a) Calculate the time taken for the droplet to reach its maximum height.
(b) Calculate the time taken for the droplet to fall from maximum height to sea level.

For (a) i just used time = distance / speed -- 56/19.8 = 2.83 seconds - is that right?

For (b) i'm a bit stuck. I assume that acceleration due to gravity will come into it. I dfidn't think i could use 'v = u + at' as it doesn't take height into consideration. I couldn't use 'v^2 = u^2 + 2as' as i didn't know what to use for 'v'. I tried using 'd = ut + 1/2 at^2' by changing it to '(d-u) / 1/2 a = t' but that came out as 11.4 secons and i don't think that's right. What do i do?

Question 2:

"Most electrical showers draw cold water direct from the main supply and heat it as it is used - day or night.

(a) Write a word equation to describe the energy transfer that takes place in an electric shower.
(b) Rewrite the equation using appropriate formula.

Any help is much appreciated.
cant be bothered to read 1 but 2 i can so
2a. electrical to heat
b. Itv=cmT

I-current
t- time
v- voltage
c- specific heat capacity
m-mass
T - change in temp
5. Ahh. specific heat capacity - forgot about that - cheers.
6. (Original post by need_help)
Just started a bit of AS revision and have no text book or good notes - got stuck on a couple of past paper questions - any ideas?

Question:

A floating fire tender is testing its pumps. The hose in use is being directed virtically upwards but the water is being blown to the left by a strong sideways wind.

Considering the path of one droplet of water:

h1 - the height above sea level to the outlet of the hose = 20m
h2 - the maximum height above sea level of the droplet = 56m

Vertical velocity of the water leaving the hose = 19.8 ms-1

By considering the vertical motion of the droplet and ignoring air resistance,

(a) Calculate the time taken for the droplet to reach its maximum height.
(b) Calculate the time taken for the droplet to fall from maximum height to sea level.

For (a) i just used time = distance / speed -- 56/19.8 = 2.83 seconds - is that right?

For (b) i'm a bit stuck. I assume that acceleration due to gravity will come into it. I dfidn't think i could use 'v = u + at' as it doesn't take height into consideration. I couldn't use 'v^2 = u^2 + 2as' as i didn't know what to use for 'v'. I tried using 'd = ut + 1/2 at^2' by changing it to '(d-u) / 1/2 a = t' but that came out as 11.4 secons and i don't think that's right. What do i do?
q1a. no that is not right as there is gravity pulling water downwards...

s=ut+.5t^2 s travelled=56-20=36
36=19.8t-4.9t^2
so t= (19.8+-sqrt(19.8^2-4*36*4.9))/9.8

but then u r doing the square root of a -ve answer which cannot be right
7. (Original post by need_help)
Just started a bit of AS revision and have no text book or good notes - got stuck on a couple of past paper questions - any ideas?

Question:

A floating fire tender is testing its pumps. The hose in use is being directed virtically upwards but the water is being blown to the left by a strong sideways wind.

Considering the path of one droplet of water:

h1 - the height above sea level to the outlet of the hose = 20m
h2 - the maximum height above sea level of the droplet = 56m

Vertical velocity of the water leaving the hose = 19.8 ms-1

(a) Calculate the time taken for the droplet to reach its maximum height.
(b) Calculate the time taken for the droplet to fall from maximum height to sea level.
That seems wrong... I calculate that the droplet will not reach that height because of gravity. Weird.

Anyway, have you done Netwon's equations of motion? THey will help you here.
8. (Original post by need_help)
edexcel - it's a specimen paper
Thanks, it was just none of this looked familliar, I was checking it wasn't my specification.
9. (Original post by mik1a)
That seems wrong... I calculate that the droplet will not reach that height because of gravity. Weird.

Anyway, have you done Netwon's equations of motion? THey will help you here.
Crap - you are right. I've just re-done the vector calculations - i messed up the scale. It should be h1 = 10m and h2 = 28m. Does that seem better. Still doesn't help me with the time questions.
10. yes that does help
s=ut+.5t^2 s travelled=28-10=18
18=19.8t-4.9t^2
so t= (19.8+-sqrt(19.8^2-4*18*4.9))/9.8
=2.66s or 1.38s...lol i think in order for there to be one solution, b^2-4ac=0
so the initital speed squared must equal 4*acceleration*h1 so u must have got sumthing else rwrong
11. i'm confused. h1 is definately 10m and h2 is 28m - just done it 3 times using differnet methods.

The next questions tells you to prove the vertical velocity of the water leaving the hose is approx 19ms-1. Originally i had used 'v^2 = u^2 + 2as' and said that:

u = 0
a = -9.81
s = 20 m (h1 - height above sea level of the outlet of the hose)

That gave v^2 as 392.4 which was equal to 19.8ms-1. Now i realise i got h1 wrong so this method doesn't really work unless i use 's' as h2-h1 = 28-10 = 18m. Using this V works out as 18.8 ms-1.

I'm still stuck on the time questions. I don't think i could have got anything else wrong.
12. (Original post by need_help)
i'm confused. h1 is definately 10m and h2 is 28m - just done it 3 times using differnet methods.

The next questions tells you to prove the vertical velocity of the water leaving the hose is approx 19ms-1. Originally i had used 'v^2 = u^2 + 2as' and said that:

u = 0
a = -9.81
s = 20 m (h1 - height above sea level of the outlet of the hose)

That gave v^2 as 392.4 which was equal to 19.8ms-1. Now i realise i got h1 wrong so this method doesn't really work unless i use 's' as h2-h1 = 28-10 = 18m. Using this V works out as 18.8 ms-1.

I'm still stuck on the time questions. I don't think i could have got anything else wrong.
r u sure that u=0
13. (Original post by lgs98jonee)
r u sure that u=0
The question doesn't give any more information so if the initial velocity isn't zero, then what else could it be?
14. well the final speed of the water when it is decleratong should not be higher than its initial velocity
15. (Original post by lgs98jonee)
well the final speed of the water when it is decleratong should not be higher than its initial velocity
Something can't be right - AS questions shouldn't be this hard. I'd go and find out from school if it wasn't half term.

Do you not think that my initial method could be right;

"For (a) i just used time = distance / speed -- 18/18.8 = 0.96 seconds

For (b) i'm a bit stuck. I assume that acceleration due to gravity will come into it. I didn't think i could use 'v = u + at' as it doesn't take height into consideration. I couldn't use 'v^2 = u^2 + 2as' as i didn't know what to use for 'v'. I tried using 'd = ut + 1/2 at^2' by changing it to '(d-u) / 1/2 a = t' but that came out as 5.71 secons and i don't think that's right."
16. On the way down it is travelling 10m further so that could account for the higher time
17. (Original post by need_help)
Something can't be right - AS questions shouldn't be this hard. I'd go and find out from school if it wasn't half term.

Do you not think that my initial method could be right;

"For (a) i just used time = distance / speed -- 18/18.8 = 0.96 seconds

For (b) i'm a bit stuck. I assume that acceleration due to gravity will come into it. I didn't think i could use 'v = u + at' as it doesn't take height into consideration. I couldn't use 'v^2 = u^2 + 2as' as i didn't know what to use for 'v'. I tried using 'd = ut + 1/2 at^2' by changing it to '(d-u) / 1/2 a = t' but that came out as 5.71 secons and i don't think that's right."
no way that a is right
for unit 1 physics u should be able to use

s=ut+(1/2)at^2 as obviously gravity will be slowing water down...using ur values i got 1.8 sumthign seconds or 2s...so it seems to go above ur desired height.

for b u use the equation above but with s as 0 as the water has not travelled any vertical net distance...i.e 18.8t-4.9t^2=0
so 4.9t(t-3.84s)=0
so t=0s or 3.84s i.e. 0s when water started at s=om and so ur answer is 3.84s

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