Solve these Geyser and Pendulum Problems with Expert Help!"

  • Thread starter Thread starter Blubla
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around two physics problems related to kinematics and energy conservation. The first problem involves calculating the initial velocity of water from a geyser that reaches a height of 40.0 m. The second problem concerns determining the speed of a pendulum at its lowest point after being released from a height of 1.5 cm.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy principles to both problems. There is a request for the original poster to show their work and clarify where they encountered difficulties.

Discussion Status

Some participants have provided feedback on the original poster's calculations, indicating that the solutions appear correct. However, there is an ongoing exploration of the methods used to arrive at those answers, with an emphasis on understanding the reasoning behind the calculations.

Contextual Notes

The original poster expresses uncertainty about solving problems that provide limited information, and there are constraints regarding posting links or images due to forum rules.

Blubla
Messages
5
Reaction score
0
Hello all,

I have a little problem solving these 2 problems, guess you all can help me because its piece

of cake for you all.

Ok here we go:

1. Old faithful geyser in yellowstone national park shoots water every hour to a height of

40.0m. With what velocity does the water leave the ground?

2. A pendulum with a mass of 1 kg is released from a height of 1.5cm above the height of

its restin position. How fast will the pendulum be moving when it passes thorugh the

lowest point of its swing?

Dunno why when I saw this kind of problem(like number 1) ,only giving 1 value, is kinda

to solve for me..

Note: Sorry about my bad english :cool:
 
Physics news on Phys.org
To get detailed help, please show your work and point out where you got stuck.

Hint: Both problems can be solved using conservation of energy.
 
Errn.. Well, I cannot post any link because I don't have 15 posts nor image because it exceeds the

required size.

I can tell the answers, 1. Vf= 28m/s

2. Vf=0.55m/s

Thats what I have at the moment, is it correct?
 
Show how you got those answers. (We don't need to see a diagram.)
 
PE = KE

vf^2= 2mgh/m=2gh

Vf^2=2*40*9.81

Vf^2=784.8

Vf=28.0m/s

----------------

PE = KE

I changed 15cm into m so it will be = 0.15

PE=mgh

PE=1(9,81)(0.15)

PE=1.5

KE=1/2mgv^2

KE= 1/2(1)

KE= 0.5

KE/PE= Vf^2

0.5/1.5=Vf^2

0.3=Vf^2(Apply square root to both side)

0.55=Vf


I got it this way.. tell me if Iam wrong
 
In the future, try to use a little more descriptive title. Perhaps something that actually pertains to the content of your inquiry.
 
Ok I will
 
Your solutions look good to me.
 

Similar threads

Replies
5
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 2 ·
Replies
2
Views
7K
Replies
2
Views
23K
  • · Replies 2 ·
Replies
2
Views
2K