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3 years ago

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4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2

5 cm cuando se tensa?

1 answer:

bazaltina [42]3 years ago

8 0

Answer:

$1030.83\ \text{N}$

Explanation:

$\Delta L$ = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = $\dfrac{\pi}{4}d^2$

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

$\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}$

La tensión en la cuerda es $1030.84\ \text{N}$ .

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A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

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Answer:

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Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

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Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

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$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

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Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

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Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

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Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

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