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krek1111 [17]
3 years ago
6

It took 1700 J of work to stretch a spring from its natural length of 2 m to a length of 6 m. Find the spring constant k.

Physics
1 answer:
algol133 years ago
4 0

Answer:

212.5 N/m

Explanation:

Work, W = 1700 J

y1 = 2 m

y2 = 6 m

y = y2 - y1 = 6 - 2 = 4 m

Let K be the spring constant.

Work done is given by

W = 1/2 Ky²

1700 = 0.5 x K x 4 x 4

K = 212.5 N/m

Thus, the spring constant is given by 212.5 N/m.

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Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.
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Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

F=\frac{k.q.Q}{r^{2}}

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}

F_{12}=536.02.10^{-3} N

Force between 2 and 3:

F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}

F_{23}=2711.63.10^{-3} N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

F_{T}=2711.63.10^{-3}-536.02.10^{-3}

F_{T}=2175.61.10^{-3} N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, F_{13}=F_{23}. Suppose distance from 1 to 3 is x, then from 2 to 3 is x-0.301

Calculating:

\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}

\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}

\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}

x^{2}=0.67x^{2}-0.40x+0.061

0.33x^{2}+0.40x-0.061=0

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

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A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2
Pepsi [2]

Work needed = 23,520 J

<h3> Further explanation </h3>

Given

height = 12 m

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work needed by the crane

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W = F x d  

W = Work, J, Nm  

F = Force, N  

d = distance, m  

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Sound waves travel fastest through a A) gas. B) liquid. C) solid. D) vacuum.
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Sound waves travel faster through <em>solids</em> than they do through gases or liquids.  <em>(C)  </em>They don't travel through vacuum at all.

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3 years ago
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