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Nostrana [21]
3 years ago
10

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?
Physics
1 answer:
Aleonysh [2.5K]3 years ago
6 0

Answer:

<em> The elastic potential energy stored in the bungee cord = 20 J</em>

Explanation:

potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below

E = 1/2ke²................ Equation 1

Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension

<em>Given: K = 10 N/m, e = 2.00 m</em>

<em>Substituting these values into Equation 1</em>

<em>E = 1/2(10)(2)²</em>

<em>E = 5×4</em>

<em>E = 20 Joules.</em>

<em>Therefore the elastic potential energy stored in the bungee cord = 20 J</em>

<em></em>

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The equation of a transverse wave traveling along a string is 1 1 y (2.00 mm)sin[(20 m )x (600 s )t] − − = − Find the (a) amplit
Lelu [443]

Answer:

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V=  30 m/s  ( + x direction )

d)  λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

As we know that standard form of wave equation given as

y=A sin(\phi -\omega t)

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s                     (2πf=ω)

ω= 2πf

f= ω /2π

f= 300/π = 95.49 Hz

K= 20 rad/m

So velocity,V

V= ω /K

V= 600 /20 = 30 m/s  ( + x direction )

V = f λ

30 = 95.49 x  λ

 λ = 0.31 m

We know that speed is the rate of displacement

U=\dfrac{dy}{dt}

U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s

8 0
3 years ago
4. A cinder block is sitting on a platform 20 m high. It has a mass of 4 kg. The block has energy. Calculate it.
zavuch27 [327]

Answer:

Explanation:

Since the block is at rest in an elevated position, we can assume that it only has potential energy.

U=mgh is the formula for potential energy where U=potential energy, m= mass, g=acceleration due to gravity, and h=height.

Plug in known variables....

U=4kg*9.8m/s^2*20m

U=784 joules of potential energy or letter A.

4 0
2 years ago
A bus hits a bug and the bug splatters on the windshield, which force is greater?
grandymaker [24]

The forces acting on a body and the type of motion that results are given by Newton's three Laws of motion

  • The <u>force </u>of the bus is <u>the same</u> as reactive force of the bug

Reason:

According to Newton's third Law of motion, given that the bug collides

with the bus, the force with which the bus hits the bug is equal to the

reactive force of the bug on the on the bus

According to Newton's second Law of motion, force is equal to the rate of

change of the momentum produced

The impulse of the force of the bus on the bug is given as follows;

F·Δt = m·(v₂ - v₁)

Given that the force of the bus is large, the change in momentum, m·(v₂ - v₁),

is also large such that the parts of the bug are split by the rapid change in

velocity, and the bug splatters on the windshield, and is then carried along

on the trip,

The equally large reactive force of the bug, is such that the bug splatters

due its magnitude

Therefore, the correct response is that <u>the forces are the same</u>

Lean more here:

brainly.com/question/21279060

8 0
2 years ago
An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 25 m lower vert
Musya8 [376]

Answer:

x = 5.79 m

Explanation:

given,

mass of the car = 39000 Kg

spring constant = 5.7 x 10⁵ N/m

acceleration due to gravity = 9.8 m/s²

height of the track = 25 m

length of spring compressed = ?

using conservation of energy

potential energy is converted into spring energy

m g h = \dfrac{1}{2}kx^2

x =\sqrt{\dfrac{2 m g h}{k}}

x =\sqrt{\dfrac{2\times 39000 \times 9.8 \times 25}{5.7 \times 10^{5}}}

x =\sqrt{33.5263}

x = 5.79 m

the spring is compressed to x = 5.79 m to stop the car.

3 0
3 years ago
A rocket is launched from the surface of Earth with a speed v0 that will allow the rocket to escape the gravitational field of E
weeeeeb [17]

Answer:

option ( a ) is correct .

Explanation:

Escape velocity on the earth = √ ( 2 GM / R )

where G is universal gravitational constant , M is mass of the earth and R is radius .

V₀ = √ ( 2 GM / R )

escape velocity on the planet where mass is equal is earth's mass and radius is 4 times that of the earth

Radius of the planet = 4 R

escape velocity of planet = √ ( 2 GM / 4R )

= .5 x √ ( 2 GM / R )

= .5 V₀

option ( a ) is correct .

8 0
3 years ago
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