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3 years ago

La osteoporosis puede presentarse en la epoca de el embaraso y la lactancia

Engineering

1 answer:

MArishka [77]3 years ago

5 0

Answer:

En algunos casos, las mujeres desarrollan osteoporosis durante el embarazo o la lactancia, aunque esto es poco común. La osteoporosis es la pérdida ósea que es lo suficientemente grave como para provocar huesos frágiles y un mayor riesgo de fractura.

Explanation:

Espero que esto ayude a marcar el MÁS CEREBRAL !!!

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pav-90 [236]

Answer:k

Explanation:

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2 years ago

Describe, in a general form, the equation, in time domain, that tells the voltage across a inductor, L, as a function of time wh

love history [14]

Answer:

a) V(t) = Ldi(t)/dt

b) If current is constant, V = 0

Explanation:

a) The voltage, V(t), across an inductor is proportional to the rate of change of the current flowing across it with time.

If V represents the Voltage across the inductor

and i(t) represents the current across the inductor in time, t.

V(t) ∝ di(t)/dt

Introducing a proportionality constant,L, which is the inductance of the inductor

The general equation describing the voltage across the inductor of inductance, L, as a function of time when a current flows through it is shown below.

V(t) = Ldi(t)/dt ..................................................(1)

b) If the current flowing through the inductor is constant i.e. does not vary with time

di(t)/dt = 0 and hence the general equation (1) above becomes

V(t) = 0

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3 years ago

The level of water in a dam is 6 m. The rectangular gate ABC is pinned at point B so it can rotate freely about this point. When

olchik [2.2K]

Answer:

The reaction at support B

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The reaction at support C

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3 years ago

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Ymorist [56]

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2 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

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3 years ago