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3 years ago

What is mechanical energy? Give an example,

Mathematics

1 answer:

Inessa [10]3 years ago

5 0

Answer:

mechanical energy ✨ is a form of energy it is all the energy that an object has because of it motion

Step-by-step explanation:

like an bow and arrow or hitting a hammer on a nail

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BRAINLIESTTT ASAP! PLEASE HELP ME :)

bonufazy [111]

Step-by-step explanation:

$sin(360 \degree - \theta) \\ = sin \: 360 \degree \: cos \: \theta - cos \: 360 \degree \: sin \: \theta \\ = 0 \times \: cos \: \theta - 1 \times sin \: \theta \\ = 0 - sin \: \theta \\ = - sin \: \theta \\ \red{ \boxed{\bold{ \therefore \: sin(360 \degree - \theta) = - sin \: \theta}}}\\$

7 0

3 years ago

Read 2 more answers

Integral of 17/(x^3-125)

daser333 [38]

Answer:

17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

Step-by-step explanation:

∫ 17 / (x³ − 125) dx

= 17 ∫ 1 / (x³ − 125) dx

= 17 ∫ 1 / ((x − 5) (x² + 5x + 25)) dx

Use partial fraction decomposition:

= 17 ∫ [ A / (x − 5) + (Bx + C) / (x² + 5x + 25) ] dx

Use common denominator to find the missing coefficients.

A (x² + 5x + 25) + (Bx + C) (x − 5) = 1

Ax² + 5Ax + 25A + Bx² − 5Bx + Cx − 5C = 1

(A + B) x² + (5A − 5B + C) x + 25A − 5C = 1

Match the coefficients and solve the system of equations.

A + B = 0

5A − 5B + C = 0

25A − 5C = 1

A = 1/75

B = -1/75

C = -2/15

So the integral is:

= 17 ∫ [ 1/75 / (x − 5) + (-1/75 x − 2/15) / (x² + 5x + 25) ] dx

Simplify:

= 17/75 ∫ [ 1 / (x − 5) − (x + 10) / (x² + 5x + 25) ] dx

Factor ½ from the numerator of the second fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 20) / (x² + 5x + 25) ] dx

Split the fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − ½ (15) / (x² + 5x + 25) ] dx

Multiply the last fraction by 4/4:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 30 / (4x² + 20x + 100) ] dx

Complete the square:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 15 / ((2x + 5)² + 75) ] dx

Split the integral:

= 17/75 ∫ 1 / (x − 5) dx − 17/150 ∫ (2x + 5) / (x² + 5x + 25) dx − 17/5 ∫ 1 / ((2x + 5)² + 75) dx

The first integral is:

∫ 1 / (x − 5) dx = ln│x − 5│

The second integral is:

∫ (2x + 5) / (x² + 5x + 25) dx = ln(x² + 5x + 25)

The third integral is:

∫ 1 / ((2x + 5)² + 75) dx = 1/√75 tan⁻¹((2x + 5) / √75)

Plug in:

= 17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

4 0

3 years ago

How many 3 digit multiples of both 4 and 6 are there?

Daniel [21]

Answer: There 225 3-digit numbers multiple of 4

5 0

3 years ago

20×(5×(-16) ) = (20×5)×(-16)

stich3 [128]

Answer:

20×(5×(-16) = (20×5)×(-16)

Step-by-step explanation:

They both equal -1600

7 0

3 years ago

Read 2 more answers

Prove that it is impossible to dissect a cube into finitely many cubes, no two of which are the same size.

solniwko [45]

explanation:

The sides of a cube are squares, and they are covered by the respective sides of the cubes covering that side of the big cube. If we can show that a sqaure cannot be descomposed in squares of different sides, then we are done.

We cover the bottom side of that square with the bottom side of smaller squares. Above each square there is at least one square. Those squares have different heights, and they can have more or less (but not equal) height than the square they have below.

There is one square, lets call it A, that has minimum height among the squares that cover the bottom line, a bigger sqaure cannot fit above A because it would overlap with A's neighbours, so the selected square, lets call it B, should have less height than A itself.

There should be a 'hole' between B and at least one of A's neighbours, this hole is a rectangle with height equal to B's height. Since we cant use squares of similar sizes, we need at least 2 squares covering the 'hole', or a big sqaure that will form another hole above B, making this problem inifnite. If we use 2 or more squares, those sqaures height's combined should be at least equal than the height of B. Lets call C the small square that is next to B and above A in the 'hole'. C has even less height than B (otherwise, C would form the 'hole' above B as we described before). There are 2 possibilities:

C has similar size than the difference between A and B
C has smaller size than the difference between A and B

If the second case would be true, next to C and above A there should be another 'hole', making this problem infinite. Assuming the first case is true, then C would fit perfectly above A and between B and A's neighborhood. Leaving a small rectangle above it that was part of the original hole.

That small rectangle has base length similar than the sides of C, so it cant be covered by a single square. The small sqaure you would use to cover that rectangle that is above to C and next to B, lets call it D, would leave another 'hole' above C and between D and A's neighborhood.

As you can see, this problem recursively forces you to use smaller and smaller squares, to a never end. You cant cover a sqaure with a finite number of squares and, as a result, you cant cover a cube with finite cubes.

3 0

4 years ago