To answer the question above, we will have to use some calculus approach to this problem.

volume of sphere = (4/3)(pi)(radius^3) = (4/3)(pi)(9^3) = 972pi 
mass = (density)(volume) = (972pi)(1,000) = 972,000pi 
force = (mass)(acceleration or gravity here) = (972,000pi)(9.8) = 9,525,600pi 
work = (force)(distance) = (9,525,600pi)(4.5) = 42,865,200pi J

I hope my answer helped you with your problem

6 0

2 years ago

Determine the correct set up for solving the equation using the quadratic formula 4x^2-5=3x+4

slava [35]

You need to subtract everything to the left side and set it equal to zero. Combine like terms.

Then, the coefficient of x^2 is a, the coefficient of x is b, and the constant term is c.

4x^2 - 5 = 3x + 4

4x^2 - 3x - 5 - 4 = 0

4x^2 - 3x - 9 = 0

a = 4; b = -3; c = -9

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(4)(-9)}}{2(4)}$

7 0

3 years ago

Complete the identity

spin [16.1K]

Answer: $cos(\pi-x)=-cos(x)$

Step-by-step explanation:

We need to apply the following identity:

$cos(A - B) = cos A*cos B + sinA*sin B$

Then, applying this, you know that for $cos(\pi-x)$ :

$cos(\pi-x)=cos(\pi)*cos(x)+sin(\pi)*sin(x)$

We need to remember that:

$cos(\pi)=-1$ and $sin(\pi)=0$

Therefore, we need to substitute these values into $cos(\pi-x)=cos(\pi)*cos(x)+sin(\pi)*sin(x)$ .

Then, you get:

$cos(\pi-x)=(-1)*cos(x)+0*sin(x)$

$cos(\pi-x)=-1cos(x)+0$

$cos(\pi-x)=-cos(x)$

8 0

2 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

3 0

2 years ago