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2 years ago

In which is momentum conserved: an elastic collision or an inelastic collision?

1 answer:

6 0

Energy and momentum<span> are always </span>conserved<span>. Kinetic energy is not </span>conserved<span> in an inelastic collision, but that is because it is converted to another form of energy (heat, etc.). The sum of all types of energy (including kinetic) is the same before and after the collision.</span>

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The answer is C. Natural gas because both A and B are their own forms of energy.

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2 years ago

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If two identical trees are cut down, one with a hand saw, and one with an electric saw...

nataly862011 [7]

The hand saw would involve more work because it takes more time and effort.

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3 years ago

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3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

$v=\sqrt{\frac{TL}{M}}$ (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

$v= \lambda f$ (2)

We can equate expression (1) and (2):

$\sqrt{\frac{TL}{M}}$ = $\lambda f$

Solving for T

$T= \frac{M(\lambda f)^2}{L}$ (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic $N_{harmonic}$ is:

$\lambda_{harmonic}=\frac{2l}{N_{harmonic}}$

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

$\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m$

We can now find T on (3) using all the values we have:

$T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}$

$T=75.22 N$

3 0

3 years ago