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Serggg [28]
2 years ago
10

suppose a 51 kg bungee jumper steps off the royal gorge bridge, in colorado. The bridge is situated 321 m above the arkansas riv

er. the bungee cords spring constant is 32 N/m, the cords relaxed length is 104m and its length is 179m when the jumpers stops falling. what is the total potential energy associated with the jumper at the end of his fall? assume that the bungee cord has negligible mass
Physics
1 answer:
Akimi4 [234]2 years ago
6 0

The total potential energy associated with the jumper at the end of his fall is 90,000 J.

The given parameters;

  • <em>mass of the jumper, m = 51 kg</em>
  • <em>height of the bridge. h  = 321 m</em>
  • <em>spring constant of the cord, k = 32 N/m</em>
  • <em>extension of the cord, x = 179 m - 104 m = 75 m</em>

The total potential energy associated with the jumper at the end of his fall is calculated as follows;

U = ¹/₂kx² + mgΔh

where;

<em>Δh is the change in height after falling </em>

U = ¹/₂(32)(75)²  + (51)(9.8)(0)

U = 90,000 J

Thus, the total potential energy associated with the jumper at the end of his fall is 90,000 J.

Learn more here:brainly.com/question/15731149

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

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3 years ago
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. A 13-g goldfinch has a speed of 8.5 m/s. What is its kinetic energy?
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PLEASE PRESS THE “Thanks!” BUTTON! :)
13 g —> 0.013 kg
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