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Kipish [7]
3 years ago
11

Elias stretches a rubber band and let's it go. The rubber band flies across the room. Elias says this demonstrates the transform

ation of kinetic energy to elastic potential energy. Is Elias correct? Explain.

Physics
1 answer:
notka56 [123]3 years ago
7 0

Answer:

This is not correct, actually it shows the transformation of elastic potential energy into kinetic energy.

Explanation:

We will explain this with and example and some numerical values to demostrate this energy transformation.

The elastic band has a behavior similar to a spring which is compressed and when released it transforms its energy from elastic to kinetic, we're going to choose a system which consists of a body 1 [kg], which is compressed against a spring with a constant of 20[kN/m], the maximum point where it is compressed is taken as the reference point where the energy potential is zero. When the package is released determine the maximum height the package reaches when released. The spring has been compressed 50 [mm].

In the attached image we can see a sketch of the mass and the spring, and the conditions of the block at maximum height.

The energy when we compress the spring will be:

U_{1-2} =\frac{1}{2} *k*dx\\where\\k=spring constant [kN/m]\\dx=spring change [m]

The energy when the block is in the highest position will be defined only by the potential energy with respect to the reference point. In this point the velocity is zero.

U_{1-2} =\frac{1}{2} *(20000)*(50*10^-3)\\U_{1-2}= 500[J]\\Elastic energy = potential energy\\U_{1-2}=Ep\\Ep=m*g*h\\500 = 1*9.81*h\\h= 50 [m]

We can see that the mass goes 50 [m] in the vertical direction, as a result of the transformation of elasticity energy into potential.

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A certain 60.0 Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 11.6 kV/m.
yuradex [85]

Explanation:

Given that,

Frequency of the power line, f = 6 Hz

Value of maximum electric field strength of 11.6 kV/m

(a) The wavelength of this very low frequency electromagnetic wave is given by using relation as :

c=f\lambda

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8\ m/s}{60\ Hz}

\lambda=5\times 10^6\ m

(b) As its can be seen that the wavelength of this wave is very high. It shows that it is a radio wave.

(c) The relation between the maximum magnetic field strength and maximum electric field strength is given by :

B_0=\dfrac{E_0}{c}\\\\B_0=\dfrac{11.6\times 10^3}{3\times 10^8}\\\\B_0=3.86\times 10^{-5}\ T

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7 0
3 years ago
Read 2 more answers
Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro
Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

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lys-0071 [83]

Answer:

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