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3 years ago

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An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break

when the elevator is at a height h = 12.5 m above the top of the spring, calculate the value the spring constant k should have so that passengers undergo a maximum acceleration of 5.3g. Take the mass of the elevators plus passengers to be M = 1439 kg.

1 answer:

Anuta_ua [19.1K]3 years ago

6 0

Answer:

$k = 9876 N/m$

Explanation:

As per energy conservation we know that

initial total gravitational potential energy = final spring potential energy

so we have

$mg(h + x) = \frac{1}{2}kx^2$

also we know that maximum acceleration will be 5.3 g

so it is given as

$a = \frac{k}{m} x$

so we have

$x = \frac{ma}{k} = \frac{5.3 mg}{k}$

$mg(h + \frac{5.3 mg}{k}) = \frac{1}{2}k(\frac{5.3mg}{k})^2$

$mg(h + \frac{5.3 mg}{k}) = \frac{14.045(mg)^2}{k}$

$h + \frac{5.3mg}{k} = \frac{14.045 mg}{k}$

$h = \frac{8.745mg}{k}$

$k = \frac{8.745 (1439)(9.81)}{12.5}$

$k = 9876 N/m$

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I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

$F = G\frac{Mm}{r^{2} }$

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = $\frac{1}{4}$ G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

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3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

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<h3>Answer</h3>

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A = cross sectional area

So it can be seen that resistance depends upon 3 factors that are length of wire , resistivity of wire and the cross sectional area of the wire.

If two of the factors, resistivity and cross sectional area, are kept constant then the resistance is directly proportional to the length of wire.

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This means that the resistance of the wire increases with the increase in length of the wire and decreases with the decrease of length of the wire.

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