# Physics AS, Urgent doubt, help needed! Watch

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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)

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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#2

(Original post by

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)

**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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(Original post by

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.

**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.

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**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.

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#5

(Original post by

Any Idea for the 2nd question as to which equation to use?

**userxx990**)Any Idea for the 2nd question as to which equation to use?

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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(Original post by

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

**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.
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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

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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(Original post by

How far have you got with it yourself?

**Stonebridge**)How far have you got with it yourself?

ive attached my working if u like.

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#10

If it's above B for 2.4 seconds it means it took

Final velocity is zero at the top.

**1.2s**up and 1.2s down.Final velocity is zero at the top.

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(Original post by

If it's above B for 2.4 seconds it means it took

Final velocity is zero at the top.

**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.

thanks alot mate!

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**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.

**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.

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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#13

It's always area

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.)

**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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(Original post by

It's always area

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.)

**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.)

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**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.)

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**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.)

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#17

There are a number of ways of doing this, including using the areas of triangles and trapeziums.

Try this algebraic method.

Distance travelled s

time is t

second section (uniform velocity)

s = vt

time is t

third section (uniform acceleration) using suvat as before

time is t

distance is s

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 t

This gives you two simultaneous equations that can be solved for t

Can you get that far?

Try this algebraic method.

Distance travelled s

_{1}in the first section (deceleration) using suvattime is t

_{1}second section (uniform velocity)

s = vt

time is t

_{2}third section (uniform acceleration) using suvat as before

time is t

_{1}as in the 1st sectiondistance is s

_{1}as in the 1st sectionthe 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 t

_{1}+ t_{2}+ t_{1}= total time which also given.This gives you two simultaneous equations that can be solved for t

_{1}and t_{2}Can you get that far?

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**Stonebridge**)

There are a number of ways of doing this, including using the areas of triangles and trapeziums.

Try this algebraic method.

Distance travelled s

_{1}in the first section (deceleration) using suvat

time is t

_{1}

second section (uniform velocity)

s = vt

time is t

_{2}

third section (uniform acceleration) using suvat as before

time is t

_{1}as in the 1st section

distance is s

_{1}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 t

_{1}+ t

_{2}+ t

_{1}= total time which also given.

This gives you two simultaneous equations that can be solved for t

_{1}and t

_{2}

Can you get that far?

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reply

**Stonebridge**)

There are a number of ways of doing this, including using the areas of triangles and trapeziums.

Try this algebraic method.

Distance travelled s

_{1}in the first section (deceleration) using suvat

time is t

_{1}

second section (uniform velocity)

s = vt

time is t

_{2}

third section (uniform acceleration) using suvat as before

time is t

_{1}as in the 1st section

distance is s

_{1}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 t

_{1}+ t

_{2}+ t

_{1}= total time which also given.

This gives you two simultaneous equations that can be solved for t

_{1}and t

_{2}

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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#20

The times for t

The distances s

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 t

t

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

Middle section

Total distance is 2s

Total time is 2t

Write the equation that adds the three distances (contains t

Write the equation that adds the times (t

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

_{1}and t_{3}are the same.The distances s

_{1}and s_{3}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 t

_{3}and s_{3}t

_{3}= t_{1}and s_{3}=s_{1}1st section (assuming your values of 30 and 10 are correct as I haven't checked this)

Middle section

Total distance is 2s

_{1}+s_{2}(because s3 = s1)Total time is 2t

_{1}+ t_{2 }(because t3 = t1)Write the equation that adds the three distances (contains t

_{1}and t_{2}as unknowns)Write the equation that adds the times (t

_{1}and t_{2}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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