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NNADVOKAT [17]
3 years ago
12

A spider of mass mm is swinging back and forth at the end of a strand of silk of length LL. During the spider's swing the strand

makes a maximum angle of θθ with the vertical. What is the speed of the spider at the low point of its motion, when the strand of silk is vertical? Express your answer in terms of the variables mmm, θθtheta, LLL, and ggg.
Physics
1 answer:
krok68 [10]3 years ago
4 0

Answer:

The speed of the spider is v = (2g*L*(1-cosθ))^1/2

Explanation:

using the energy conservation equation we have to:

Ek1 + Ep1 = Ek2 + Ep2

where

Ek1 = kinetic energy = 0

Ep1 = potential energy = m*g*L*cosθ

Ek2 = (m*v^2)/2

Ep2 = m*g*L

Replacing, we have:

0 - m*g*L*cosθ = (m*v^2)/2 - m*g*L

(m*v^2)/2 = m*g*L*(1-cosθ)

v^2 = 2g*L*(1-cosθ)

v = (2g*L*(1-cosθ))^1/2

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The distance between adjacent nodes in a standing wave pattern in a length of string is 25.0 cm:A. What is the wavelength of wav
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A) 50 cm

B) 10000 cm/s

Explanation

Step 1

A)

If you know the distance between nodes and antinodes then use this equation:

\begin{gathered} \frac{\lambda}{2}=D \\ \text{where}\lambda\text{ is the wavelength} \\ D\text{ is the distance betw}een\text{ nodes} \end{gathered}

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replace and evaluate

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1 year ago
A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
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Answer:

0.6983 m/s

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Using conservation of energy

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m v² = k x²

(0.021) v² = (0.4) (0.16)²

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