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Ivan
3 years ago
14

How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m when the cord is stretched 2.00

m?
A. 10.0 J
B. 20.0 J
C. 40.0 J
D. 200 J
Physics
2 answers:
MrRissso [65]3 years ago
7 0

Elastic potential energy stored in bungee cord is 20.0 J.

Answer: Option B

<u>Explanation: </u>

The elastic potential energy is a form of potential energy stored in elastic materials. Generally potential energy is the energy of any object when it is at rest. But in materials possessing elastic properties, they store these potential energy in the form of elastic potential energy and it is used when the elastic materials undergoes elastic deformation.

For example, if we consider a spring, the elastic potential energy of the spring will be zero, when it is not deformed elastically i.e., when the spring is not stretched or compressed. But the same spring will be exhibiting elastic potential energy when the spring is stretched or compressed. So, the formula for elastic potential energy is

      \text { Elastic potential energy }=\frac{1}{2} \times \text { spring constant } \times\left(\frac{\text {stretching}}{\text {compressing distance}}\right)^{2}

In this case, as the spring constant of cord is given as 10 N/m and the stretching distance of the cord is given as 2 m, so the elastic potential energy stored in the cord will be

       \text { Elastic potential energy }=\frac{1}{2} \times 10 \times(2)^{2}=\frac{1}{2} \times 10 \times 4 = 20 J

UNO [17]3 years ago
5 0

the correct answer is B. 20.0 J

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A bungee jumper jumps off a bridge and bounces up and down several times.
Stella [2.4K]

The energy that was lost due to air resistance while she was bouncing is determined as 3,360 J.

<h3>Conservation of energy</h3>

The amount of energy lost due to  air resistance while she was bouncing is determined from the principle of conservation of energy.

ΔE = P.E - Ux

ΔE = mgh - ¹/₂kx²

ΔE = (50)(9.8)(16) - ¹/₂(35)(16)²

ΔE = 3,360 J

Thus, the energy that was lost due to air resistance while she was bouncing is determined as 3,360 J.

Learn more about energy here: brainly.com/question/13881533

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If the Moon's orbit were perpendicular to the ecliptic, would eclipses be possible? What would be the most common phase of the M
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If a fully loaded shopping cart has a head-on collision with an empty cart both moving at the same speed toward each other, whic
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4 years ago
A train is 240 meters long and travels 20 m/s. How long does
const2013 [10]

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=18 sec

Explanation:

240m=20m/s, find 360m

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6 0
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A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of
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We have the relation

\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}

where v_{A \mid B} denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}

v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}

Let's assume the river flows South-to-North, so that

\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath

and let -90^\circ < \theta < 90^\circ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)

Then the velocity of the boat relative to the Earth is

\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath

The crossing is 153.0 m wide, so that for some time t we have

153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s

which is minimized when \theta=0^\circ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}

The boat's position \vec x at time t is

\vec x = \vec v_{B\mid E} t

so that after 30.0 s, the boat's final position on the other side of the river is

\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath

and the boat would have traveled a total distance of

\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}

3 0
2 years ago
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