A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched by 10 cm. How far would the spring need to be stretched to give the spring 20 J of elastic potential energy? Use proportional reasoning.

Option 1: 14 cm
Option 2: 28cm
Option 3: 20 cm
Option 4: 40 cm

From the given information in the question, the correct option is Option 1: 14 cm.

A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = k

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = k

10 = k

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = k

20 = (0.2)

40 = 0.2

= 200

x =

  = 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

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i don't know, you need to give me some answers.

explanation:

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.