How do I Calculate Average Thrust when given t, v, delta m

Click For Summary
To calculate average thrust, the formula used is Thrust = v exhaust * (delta mass / time). In this case, substituting the values gives Thrust = 5000 m/s * (1.000 kg / 0.0050 s), resulting in 10,000,000 kg m/s. The impulse is calculated as Impulse = v exhaust * delta mass, yielding 5000 kg m/s. There is a suggestion to check the number of zeroes in the final answer for precision. It's also recommended to express the result in scientific notation for clarity.
BigREDgrin
Messages
2
Reaction score
1

Homework Statement


During a .0050 second time period a rocket expels 1.000kg of gas at a velocity of 5000 m/s. Calculate the rockets average Thrust, as well as Impulse. http://imgur.com/a/uE9dVI am going to use this note for reference on a test and want to rock solid about it.

2. Homework Equations

I have the formulas given by my professor are

Thrust= v exhaust * (delta mass / time)

Impulse rocket= v exhasut (delta mass)= delta p rocket

The Attempt at a Solution


[/B]
Like I said I want to be sure of this so I can refer to it later. Here is my best try

Thrust= (5000m/s)*(1.000kg/.0050s) = 5000 m/s * 200 kg/s= 10000000 kg m/s

Impulse= 5000 m/s (1.000 kg) = 5000kg m/s

So please tell me if and where I have gone wrong.
 
Last edited by a moderator:
  • Like
Likes Cameron Kerr
Physics news on Phys.org
Looks good from where I'm at.
 
Thanks for your response. I am in the process of doing the same with Coefficient of Restitution, speed of approach over speed of separation. Thanks again
 
BigREDgrin said:
10000000 kg m/s
Check the number of zeroes.
The way the information is stated makes me suspect they are also looking for the right precision being expressed in the answer. It might be better to write it as 1.000 (or whatever is the right precision) times a power of ten.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
9K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 14 ·
Replies
14
Views
8K