Physics AS, Urgent doubt, help needed! Watch

userxx990
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fgftgrtrtrt

need help!

i have 2 confusions in the fluid dynamics,

1. i knw all the forces on a ball moving downwards through a fluid, but what are the forces on the ball if it moves up through the fluid?

2. For calculating the Velocity for the ball falling/going up the fluid which formula should we use, there are 2 to use.

* v= 2*r^2*g (Ps-Pf)/9*viscosity

or

* v= F/6*pie(viscosity*radius)
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Stonebridge
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(Original post by userxx990)
fgftgrtrtrt

need help!

i have 2 confusions in the fluid dynamics,

1. i knw all the forces on a ball moving downwards through a fluid, but what are the forces on the ball if it moves up through the fluid?

2. For calculating the Velocity for the ball falling/going up the fluid which formula should we use, there are 2 to use.

* v= 2*r^2*g (Ps-Pf)/9*viscosity

or

* v= F/6*pie(viscosity*radius)


The forces would be the same. The only difference is that the viscous drag force is in the opposite direction. That is, in the downward direction, whereas it's upwards when the ball falls. Gravity and Archimedes upthrust will be the same.
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userxx990
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(Original post by Stonebridge)
The forces would be the same. The only difference is that the viscous drag force is in the opposite direction. That is, in the downward direction, whereas it's upwards when the ball falls. Gravity and Archimedes upthrust will be the same.
Thank You!
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userxx990
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(Original post by Stonebridge)
The forces would be the same. The only difference is that the viscous drag force is in the opposite direction. That is, in the downward direction, whereas it's upwards when the ball falls. Gravity and Archimedes upthrust will be the same.
Any Idea for the 2nd question as to which equation to use?
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Stonebridge
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(Original post by userxx990)
Any Idea for the 2nd question as to which equation to use?
The one with F needs a value for F, the viscous drag force.
The1st equation you have there is what you get when you sub for this force F and eliminate it.

The 1st equation gives the terminal velocity in terms of the quantities you know. That is the standard formula derived from Stokes' Law for a falling object.
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userxx990
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(Original post by Stonebridge)
The one with F needs a value for F, the viscous drag force.
The1st equation you have there is what you get when you sub for this force F and eliminate it.

The 1st equation gives the terminal velocity in terms of the quantities you know. That is the standard formula derived from Stokes' Law for a falling object.
Oh ok, so we can use both?
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userxx990
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Could u help?

A Particle P is projected vertically upwards with a speed of 30 m/s from a point A.
The Point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4 seconds.

Calculate the Value of h
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Stonebridge
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How far have you got with it yourself?
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userxx990
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(Original post by Stonebridge)
How far have you got with it yourself?
for more than an hour i was into this, but couldnt solve, but i solved some how and i got 43.7 m but the ans is 39.

ive attached my working if u like.
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Stonebridge
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If it's above B for 2.4 seconds it means it took 1.2s up and 1.2s down.
Final velocity is zero at the top.
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userxx990
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(Original post by Stonebridge)
If it's above B for 2.4 seconds it means it took 1.2s up and 1.2s down.
Final velocity is zero at the top.
Oh Yeaaah! i got 39 now.

thanks alot mate!
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userxx990
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(Original post by Stonebridge)
If it's above B for 2.4 seconds it means it took 1.2s up and 1.2s down.
Final velocity is zero at the top.
Last question, i hope so :P but please help if u can..


i have drawn the graph for u, jus tell me how to do the a. and b part.


and is the area under for this specific graph above it (shaded region) or below it?


this is the question.

The brakes of a train which is travelling at 108 KM/H Are applied as the train passes a point A. The brakes produce a retardation of magnitude 3f ms/square until the speed of the train is reduced to 36km/h
The train travels at this speed for a distance and is then uniformly accelerated at f ms/square until it again reached the speed 108KM/H as it passes point B. The time taken by the train in travelling from A to B, a distance of 4 km, is 4 Minutes.


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Stonebridge
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It's always area under the graph.
Make sure your units are consistent.
If your time axis is seconds the convert speeds to m/s
Distances (area) will all then be in m.

The value of f is the gradient of the sloping line. (The tangent of the angle it makes with the horizontal.)
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userxx990
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(Original post by Stonebridge)
It's always area under the graph.
Make sure your units are consistent.
If your time axis is seconds the convert speeds to m/s
Distances (area) will all then be in m.

The value of f is the gradient of the sloping line. (The tangent of the angle it makes with the horizontal.)
Hmm so is that a trapezium when the train is decelerating? I'm so confused if I should first calculate the time or the distance in order to find the F
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userxx990
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(Original post by Stonebridge)
It's always area under the graph.
Make sure your units are consistent.
If your time axis is seconds the convert speeds to m/s
Distances (area) will all then be in m.

The value of f is the gradient of the sloping line. (The tangent of the angle it makes with the horizontal.)
Did not get it, its very hard
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userxx990
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(Original post by Stonebridge)
It's always area under the graph.
Make sure your units are consistent.
If your time axis is seconds the convert speeds to m/s
Distances (area) will all then be in m.

The value of f is the gradient of the sloping line. (The tangent of the angle it makes with the horizontal.)
i have drawn the graph and everything but i cant seem to find out the value of f, :/
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Stonebridge
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There are a number of ways of doing this, including using the areas of triangles and trapeziums.

Try this algebraic method.

Distance travelled s1 in the first section (deceleration) using suvat

 s = \frac{(u +v)}{2} t

time is t1

second section (uniform velocity)

s = vt

time is t2

third section (uniform acceleration) using suvat as before

 s = \frac{(u +v)}{2} t

time is t1 as in the 1st section
distance is s1 as in the 1st section

the total distance travelled is given so you know the sum of the three distances.

Write down the expression for the sum of the 3 distances.

You also know that t1 + t2 + t1 = total time which also given.

This gives you two simultaneous equations that can be solved for t1 and t2

Can you get that far?
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userxx990
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(Original post by Stonebridge)
There are a number of ways of doing this, including using the areas of triangles and trapeziums.

Try this algebraic method.

Distance travelled s1 in the first section (deceleration) using suvat

 s = \frac{(u +v)}{2} t

time is t1

second section (uniform velocity)

s = vt

time is t2

third section (uniform acceleration) using suvat as before

 s = \frac{(u +v)}{2} t

time is t1 as in the 1st section
distance is s1 as in the 1st section

the total distance travelled is given so you know the sum of the three distances.

Write down the expression for the sum of the 3 distances.

You also know that t1 + t2 + t1 = total time which also given.

This gives you two simultaneous equations that can be solved for t1 and t2




Can you get that far?

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userxx990
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(Original post by Stonebridge)
There are a number of ways of doing this, including using the areas of triangles and trapeziums.

Try this algebraic method.

Distance travelled s1 in the first section (deceleration) using suvat

 s = \frac{(u +v)}{2} t

time is t1

second section (uniform velocity)

s = vt

time is t2

third section (uniform acceleration) using suvat as before

 s = \frac{(u +v)}{2} t

time is t1 as in the 1st section
distance is s1 as in the 1st section

the total distance travelled is given so you know the sum of the three distances.

Write down the expression for the sum of the 3 distances.

You also know that t1 + t2 + t1 = total time which also given.

This gives you two simultaneous equations that can be solved for t1 and t2

Can you get that far?

Yeah bro i tried all u told but i don't knw where I'm going wrong though,
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Stonebridge
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The times for t1 and t3 are the same.
The distances s1 and s3 are the same.

The deceleration and acceleration are between the same two velocities at the same rate (3f), so will take the same time and cover the same distance. So there is no need for t3 and s3
t3 = t1 and s3=s1

1st section (assuming your values of 30 and 10 are correct as I haven't checked this)

s_1 = \frac{(30+10)}{2} t_1

Middle section

s_2 = 10 t_2

Total distance is 2s1 +s2 (because s3 = s1)

Total time is 2t1 + t2 (because t3 = t1)

Write the equation that adds the three distances (contains t1 and t2 as unknowns)

Write the equation that adds the times (t1 and t2 as unknowns)

You now have two equations with two unknowns as you have values for the total distance and total time.
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