Un cuerpo de 180 kg se debe lanzar hacia una distancia de 25 m. ¿Qué valor tiene el trabajo que se realiza al lanzarlo?

Mathematics

1 answer:

dsp732 years ago

3 0

Answer:

45, 000 J o 45 KJ

Step-by-step explanation:

Recordemos que trabajo realizado = fuerza * distancia

En este caso, fuerza = peso del objeto

Por lo tanto, peso del objeto = 180 Kg * 10ms ^ 2 = 1800 N

Por eso;

Trabajo realizado = 1800 N * 25 m = 45, 000 J o 45 KJ

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What is the binomial expansion of (x 2y)7? 2x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5 14xy6 2y7 x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5

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You can take x = a, 2y = b and then can apply the binomial theorem.

The expansion of given expression is given by:

Option D: $x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7$ is

<h3>What is binomial theorem?</h3>

It provides algebraic expansion of exponentiated(integer) binomial.

According to binomial theorem,

$(a+b)^n = \sum_{i=0}^n ^nC_i a^ib^{n-i}$

<h3>How to use binomial theorem for given expression?</h3>

Taking a = x, and b =2y, we have n = 7, thus:

$(x+2y)^7 = \: ^7C_0x^0(2y)^7 + \: ^7C_1x^1(2y)^6 + \: ^7C_2x^2(2y)^5 + \: ^7C_3x^3(2y)^4 + \:^7C_4x^4(2y)^3 + \:^7C_5x^5(2y)^2 + \:^7C_6x^6y^1 + \: ^7C_0x^7y^0\\\\
(x+2y)^7 = 128y^7 + 448xy^6 + 672x^2y^5 + 560x^3y^4 + 280x^4y^3 + 84x^5y^2 + 14x^6y + x^7\\\\
(x+2y)^7 = x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7$