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Bogdan [553]
3 years ago
8

A block slides with no friction and hits a spring with spring constant k=2000 N/m. The block compresses the spring in a straight

line for a distance 0.15m. The block’s kinetic energy, in J, at that point is 0 J. What was its initial kinetic energy, in Joules?
Physics
1 answer:
tamaranim1 [39]3 years ago
5 0

To solve this exercise it is necessary to apply the concepts related to kinetic energy in a spring.

By definition the kinetic energy keeping what is stipulated in the stable Hooke law that

KE_s = \frac{1}{2} kx^2

Where,

K = Spring constant

x = Displacement

Our values are given as,

x = 0.15m

k = 2000N/m

Therefore replacing

KE_s = \frac{1}{2} kx^2

KE_s = \frac{1}{2} (2000)(0.15)^2

KE_s = 22.5J

Therefore the initial kinetic energy is 22.5J

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A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture f/4.00 to photograph a bear t
labwork [276]

Answer:

<h3>(A) The width x = 0.24 \times 10^{-3} m</h3><h3>(B) The new width is 1.32 \times 10^{-3} m</h3>

Explanation:

Given :

Focal length f =   135 \times 10^{-3}  m

Maximum aperture D = \frac{f}{4}

Wavelength \lambda = 550 \times 10^{-9} m

(A)

From rayleigh criterion,

  \theta = \frac{1.22 \lambda }{D}

  \theta =\frac{ 1.22 \times 550 \times 10^{-9}  }{33.75 \times 10^{-3} }

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From angle formula,

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x = 12 \times 1.98 \times 10^{-5} m

x = 23.76 \times 10^{-5}

x = 0.24 \times 10^{-3} m

(B)

We know that \theta is proportional to the x and inversely proportional to the D

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   x = 0.24 \times  10^{-3}  \times \frac{22}{4}

   x = 1.32 \times 10^{-3} m

6 0
3 years ago
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