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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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One clue we can observe that means a chemical reaction has occurred is the formation of light. What is one chemical reaction you
dimaraw [331]

Answer:

There are five signs of a chemical change:

Color Change.

Production of an odor.

Change of Temperature.

Evolution of a gas (formation of bubbles)

Precipitate (formation of a solid)

Explanation:

I just went ahead and gave you the five signs of chemical change hoped it helped

8 0
3 years ago
A car moves with a speed of 30 metres per second calculate the distance travelled in 30 seconds
Leni [432]

30x30=900

The answer is 900 meters after 30 seconds

7 0
3 years ago
A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec
Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 8=50.24rad/sec

Velocity of the string wave is equal to v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec

Power of wave propagation is equal to P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt

So power of the wave will be equal to 5.464 watt

6 0
3 years ago
For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. What is the
galina1969 [7]

Answer:

The first frequency of audible sound in the speed sound is

f = 662 Hz

Explanation:

vs = 344 m/s

x =  52 cm * 1 / 100m = 0.52m

The wave length is the distance between the peak and peak so

d = 2x

d = 2*0.52 m

d = 1.04 m

So the frequency in the speed velocity is

f = 1 / T

f = vs / x = 344 m/s / 0.52m

f ≅ 662 Hz

7 0
4 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
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