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3 years ago

6

un resorte de 10cm de longitud recibe una magnitud de fuerza que lo estira hasta medir 15cm ¿cual es la magnitud de la tension u

nitaria o deformacion lineal? Si me explican la fórmula estaría genial!

1 answer:

Ber [7]3 years ago

8 0

Answer: 0.5

Explanation:

The modulus of elasticity (called <em>"alargamiento unitario"</em> in spanish) $\epsilon$ of a spring is given by the following formula:

$\epsilon=\frac{\Delta L}{L}$

Where:

$L=10 cm$ is the original length of the spring

$\Delta L=L_{f}-L$ is the elongation of the spring, being $L_{f}=15 cm$ the length of the spring after a force is applied to it.

$\epsilon=\frac{L_{f}-L}{L}=\frac{15 cm - 10 cm}{10 cm}$

Then:

$\epsilon=0.5$

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To solve the problem, apply the concepts related to constructive and destructive interference. For this purpose, we will start by defining the distance from the dark fringe from the central maximum in terms of the slit separation, the number of fringe the wavelength and the distance of the screen to the slit. Mathematically this is,

$y_n = \frac{(2n-1)\lambda D}{2d}$

$D = \frac{2dy_n}{(2n-1)\lambda}$

Here,

d = Slit separation

n = Number of fringe

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D = Distance of the screen to the slit

$\lambda$ = Wavelength

Our values are given as,

$\lambda = 679nm = 679*10^{-9}m$

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Replacing,

$D = \frac{2(1.1*10^{-3})(8.01*10^{-2})}{(2(14)-1)(679*10^{-9})}$

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Therefore the screen is located from the slit to 9.61m

7 0

3 years ago

What other types of energy are present when turning a generator that were not mention in question ? Explain where in the process

klemol [59]

Answer:

When an object is in motion, there is energy associated with that object. Why should that be the case? Moving objects are capable of causing a change, or, put differently, of doing work. For example, think of a wrecking ball. Even a slow-moving wrecking ball can do a lot of damage to another object, such as an empty house. However, a wrecking ball that is not moving does not do any work

(hope it helps :p )

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3 years ago

In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin

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Answer:

Y = 0.0254 m = 25.4 mm

Explanation:

The formula for the fringe spacing in Young's Double-slit experiment is given by the following formula:

$Y = \frac{\lambda L}{d}$

where,

Y = fringe spacing = 0.0176 m

λ = wavelength = 425 nm = 4.25 x 10⁻⁷ m

L = Distance between screen and slits

d = slit separation

Therefore,

$\frac{0.0176\ m}{4.25\ x\ 10^{-7}\ m} = \frac{L}{d}\\\\\frac{L}{d} = 41411.76$

Now, for:

λ = 614 nm = 6.14 x 10⁻⁷ m

$Y = (6.14\ x\ 10^{-7}\ m)(41411.76)\\$

<u>Y = 0.0254 m = 25.4 mm</u>

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2 years ago

Utility poles are to be set every 30 meters. How many poles will be set in one mile if one was set at the beginning of the mile)

Zepler [3.9K]

Answer:

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Because there are 1609.34 meters in a mile. 1609.34÷30=53.64 but because you put one at the beginning of the mile it will stay 53 and not round up to 54

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3 years ago

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