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Leona [35]
3 years ago
8

An unstrained horizontal spring has a length of 0.31 m and a spring constant of 220 N/m. Two small charged objects are attached

to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.022 m relative to its unstrained length. Determine (
a. the possible algebraic signs and (
b. the magnitude of the charges.
Physics
1 answer:
Kruka [31]3 years ago
8 0
The solution you should use is Hooke's law: F=-kx

It should have the same signs because they repel due to the stretch of the spring. 

a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be 
<span>F = kx 
270 N/m x 0.38 m = 102.6 N 
</span>
b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force. 



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(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

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<h3>Stage 2</h3>

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<h3>y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\</h3><h3 /><h3>Stage 3</h3>

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y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2

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To increase the centripetal acceleration to 2.00 m/s^2, you can double the speed or decrease the radius by 1/4

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Learn more about centripetal acceleration:

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