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liberstina [14]
3 years ago
11

A bungee cord stretches 25 meters and has a spring constant of 140 N/m. How much energy is stored in the bungee cord

Physics
1 answer:
goldenfox [79]3 years ago
5 0

Answer:

The energy stored in the bungee cord is 43,750 J.

Explanation:

Given;

extension of the bungee cord, x = 25 m

spring constant of the bungee cord, k = 140 N/m

The elastic potential energy stored in the bungee cord is calculated as;

U = ¹/₂kx²

U = ¹/₂(140)(25)²

U = 43,750 J

Therefore, the energy stored in the bungee cord is 43,750 J.

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I believe the answer is C

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Masja [62]
The correct answer is 
<span>C) either the pressure of the gas, the volume of the gas, or both, will increase.

In fact, the ideal gas law can be written as
</span>pV=nRT
<span>where 
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3 years ago
Hurricane or typhoons occur when large areas of the open ocean soak up heat from the sun t or f​
11Alexandr11 [23.1K]

Answer: true

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3 years ago
A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan
PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d)  V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

                                 y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

                                 t =3.349 s

- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

                                 150 = 0 + V_x,i*3.349

                                  V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

                                 V_y,f = V_y,i + g*t

                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

5 0
3 years ago
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Allushta [10]

No.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.

<h3>What is a random error?</h3>

Random error is defined as the deviation of the total error from its mean value due to chance.

Random errors can result from the instrument not being precise or from mistakes by the researcher.

Random errors can be minimized by taking multiple readings and averaging the results.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a ransom error has been minimized.

Learn more about random errors at: brainly.com/question/22041172

3 0
3 years ago
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