What is the unknown air resistance co-efficient of the balloon?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the air resistance coefficient (k) for a balloon using the relationship between mass, gravitational force, and terminal velocity. Participants are exploring the equations related to air resistance and terminal velocity.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between mass, gravitational force, and terminal velocity, specifically how to rearrange the equations to isolate the air resistance coefficient (k). Questions arise regarding the correct interpretation of mass and velocity in the context of the equations provided.

Discussion Status

There is ongoing exploration of the equations and their components, with some participants attempting to clarify the definitions and values needed to calculate k. Confusion is noted regarding the values to be used for mass and terminal velocity, indicating that the discussion is still active and seeking clarity.

Contextual Notes

Participants are working with specific values for mass and gravitational acceleration but express uncertainty about the correct application of these values in the equations. The discussion reflects a need for further clarification on the definitions and assumptions involved in the calculations.

luigihs
Messages
86
Reaction score
0
1. Calculate the air resistance co-efficient of the ballonI trying to do my best because I have this equations in my notes Fd = k . v^2 and its says the force of air resistance is proportional to the square of the velocity of the object.

So I only now v^2 I think is my terminal velocity that is 2.9584 , but I don't know K because is the air resistance coefficient, and I don't know the force drag ( Fd)

NOW I have this other equation For an object with mass m and air resistance coefficient k the terminal velocity is given by:

Vt = √(m.g/k)

and this I know Vt is 2.9584 , m is 2.3192g , and g is 9.8 but the only thing I don't know is K because is the air coefficient , and I can rearrange the equation like k = √(m.g/vT) ... but I not sure please help me out guys.
 
Physics news on Phys.org
If you have to find "k":

mg = kv^2, when the object is in terminal velocity.
Rearrange we have mg/v^2 = k.

m = 2.3192 grams or 2.3192 x g ?

What is your problem actually?
 
physicsisgrea said:
If you have to find "k":

mg = kv^2, when the object is in terminal velocity.
Rearrange we have mg/v^2 = k.

m = 2.3192 grams or 2.3192 x g ?

What is your problem actually?

mg/v^2 = k

Ok so the balloon mass is 2.3192 grams , g = 9.8 m/s^2 right? , and v is that my terminal velocity or what velocity do I have to put ?
 
luigihs said:
mg/v^2 = k

Ok so the balloon mass is 2.3192 grams , g = 9.8 m/s^2 right? , and v is that my terminal velocity or what velocity do I have to put ?

luigihs said:
mg/v^2 = k

Ok so the balloon mass is 2.3192 grams , g = 9.8 m/s^2 right? , and v is that my terminal velocity or what velocity do I have to put ?

If you have find k, based on your equation "air drag = k times v^2",
k = mg/v^2 = 0.002596871681

Assume v is in m/s

Plug in the terminal velocity cause "air drag equals to mg" when it has the terminal velocity.
 
physicsisgrea said:
If you have find k, based on your equation "air drag = k times v^2",
k = mg/v^2 = 0.002596871681

Assume v is in m/s

Plug in the terminal velocity cause "air drag equals to mg" when it has the terminal velocity.

Where do you get that number sorry I am confused! :(
 
luigihs said:
Where do you get that number sorry I am confused! :(

put m = mass of the object
put g = 9.8 m/s^2
put v = terminal velocity
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K