Solve these Geyser and Pendulum Problems with Expert Help!"

  • Thread starter Blubla
  • Start date
In summary: Just make sure to always include units in your calculations and final answers for accuracy and clarity. In summary, the conversation discusses two physics problems involving the Old Faithful geyser in Yellowstone National Park and a pendulum. The first problem involves finding the velocity at which the water leaves the ground and the second problem involves finding the speed of the pendulum at the lowest point of its swing. The conversation also includes hints and solutions for solving these problems using conservation of energy. The solutions are Vf=28m/s for the first problem and Vf=0.55m/s for the second problem.
  • #1
Blubla
5
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:
 
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  • #2
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.
 
  • #3
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?
 
  • #4
Show how you got those answers. (We don't need to see a diagram.)
 
  • #5
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
 
  • #6
In the future, try to use a little more descriptive title. Perhaps something that actually pertains to the content of your inquiry.
 
  • #7
Ok I will
 
  • #8
Your solutions look good to me.
 

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