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4 years ago

15

A 20 cm long spring is attached to a wall. the spring stretches to a length of 22 cm when you pull on it with a force of 100 n.

what is the spring constant, "k" of the spring?

1 answer:

liberstina [14]4 years ago

7 0

This phenomenon can be describedby Hooke'slaw and expressed by the formula: F=k*deltax, where k is the spring's constant, and delta xis the displacement of the spring. Transposing the terms such that we want to know the value of k,

k = F/deltax = 100N/(22 cm - 20cm)
k = 50 N/cm

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3 years ago

A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 4.0\,\text m4.0m4, point, 0, start text, m, end

kirill [66]

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4 years ago

Hooke's law describes a certain light spring of unstretched length 33.6 cm. when one end is attached to the top of a doorframe a

masha68 [24]

Missing question: "What is the spring's constant?"

Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
$F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N$
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
$\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m$
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4 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

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heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

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3 years ago