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Marat540 [252]
2 years ago
14

Evaluate the given integral by using substitution method:

5sin%28%5Cfrac%7B2%5Cpi%20%7D%7Bx%7D%20%29%7D%7Bx%5E%7B2%7D%20%7D%20%7D%20%5C%2C%20dx" id="TexFormula1" title="\int\limits^a_b {\frac{5sin(\frac{2\pi }{x} )}{x^{2} } } \, dx" alt="\int\limits^a_b {\frac{5sin(\frac{2\pi }{x} )}{x^{2} } } \, dx" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Veronika [31]2 years ago
5 0

Step-by-step explanation:

substitute the angle of sine with a variable and differentiate

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A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and
goblinko [34]

Answer:

D:Spring A is 3 m longer than spring B because 15 – 12 = 3.

Step-by-step explanation:

In this question, you should remember the Hooke's Law in physics.

The Hooke's Law simply explains that the extension that occurs on a spring is directly proportional to the load applied on it.

The mathematical expression for this law is

F=-kx

where;

F= force applied on the spring

x = the extension on the spring

k= the spring constant which varies in spring.

The question will need you to calculate the extension on the springs A and B then compare the values obtained.

<u>In spring A</u>

Force, F=60N and spring constant ,k=4 N/m

To find the extension x apply the expression;

F=-kx\\\\60=-4*x\\\\60=-4x\\\\\frac{60}{-4} =\frac{-4x}{-4} \\\\\\-15=x

Here the spring extension is 15 m

<u>In spring B</u>

Force, F=60N and spring constant , k=5N/m

To find the extension x apply the same expression

F=-kx\\\\60=-5*x\\\\60=-5x\\\\\\\frac{60}{-5} =\frac{-5x}{-5} \\\\\\-12=x

Here the extension on the spring is 12 m

<u>Compare</u>

The extension on spring A is 3 m longer than that in spring B because when you subtract the value of spring B from that in spring A you get 3m

=15m-12m=3m

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Step-by-step explanation:

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2 years ago
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6       6
x = 7  

Each friend received 7 cards
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3 years ago
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