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ICE Princess25 [194]
3 years ago
5

How do you find the slope of a force-extension graph?

Physics
1 answer:
Harrizon [31]3 years ago
3 0

The relationship between force and extension is a linear one, which means that if you plot a force vs. extension graph, you'll get a straight line. It will pass through the origin (x = 0; F = 0), and its slope will be equal to the spring constant, k.

Measure the Slope of the Force Extension Graph

In general, you can find the slope of a line by choosing two points and forming a ratio of the rise and the run between these two points. If the first point you choose is (x1, F1), and the second point is (x2, F2), the slope of the line is:

slope= f(2)- f(1)

           ---------

            x(2)-x(1)

Assuming F2 is larger than F1.

This is the value of the spring constant, k. Despite the minus sign in the Hooke's law equation, k is a positive number, because the slope in the Hooke's law graph is positive.

Note that the spring constant has units of force/distance. In the MKS system, the spring constant units are newtons/meter. In the CGS system, they are dynes/centimeter. In the imperial system, they are pounds of force (lbf) /foot.

Now that you have the spring constant, you can predict exactly how much the spring will distend or compress when you subject it to any force.

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A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this
wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

r=r_o\times A^{\frac{1}{3}}

r_o=1.25 \times 10^{-15} m = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}

r=4.6656\times 10^{-15} m=4.6656 fm

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) F=k\times \frac{q_1q_2}{a^2}

k=9\times 10^9 N m^2/C^2 = Coulombs constant

q_1,q_2 = charges kept at distance 'a' from each other

F = electrostatic force between charges

q_1=+1.602\times 10^{-19} C

q_2=+1.602\times 10^{-19} C

Force of repulsion between two protons on opposite sides of the diameter

a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m

F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}

F=2.6527 N

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

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