Which on is bigger 7/8 or 5/6

1 answer:

6 0

Nope - 7/8 is greater Here's a cool way to solve these types of problems First write 5/6ths and 7/8ths on a sheet of paper. Then multiply the numbers that are across from each other. 8x5 and 7x6 are the numbers. 8x5=40 7x6=42 42 is greater than 40, and the numerator of the fraction used in the equation is the bigger fraction. 7 is the numerator of 7/8 which is bigger than 5/6.

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Suppose that F is an inverse square force field below, where c is a constant.

Andre45 [30]

Answer:

Forces in our Universe

Step-by-step explanation:

First of all we have,

$F(r) = \frac{cr}{|r|^{3} }$

and,

$r = xi+yj+zk$

We need to define a function that allows us to find said change based on r, so one of the functions that shows that change is,

$f(r) = - c /|r|$

That is,

$\nabla f = F$

For this case F is a conservative field and the line integral is independent of the path. Thus, defining $P_{1} = (x_{1}, y_{1}, z_{1})$ and $P_{2} = (x_{2}, y_{2}, z_{2})$ . So the amount of work on the movement of the object from P1 to P2 is,

$W=\int\limits_c {F} \, dr$

$W= f(P_{1}-P_{2})$

$W= \frac{c}{(x_{2}^2+ y_{2}^2+ z_{2}^2)^{1/2} } -\frac{c}{(x_{1}^2+ y_{1}^2+ z_{1}^2)^{1/2}}$

$W= c(\frac{1}{d1}-\frac{1}{d2} )$

2) The gravitational force field is given by,

$F(r) =-\frac{mMGr}{ |r|^3}$

The maximum distance from the earth to the sun is $1.52*10 ^ 8$ km and the minimum distance is $1.47*10 ^ 8$ km. The mass values of the bodies are given by m = $5.97*10 ^ {24}$ kg, M = $1.99 *19 ^ {30}$ kg and the constant G is $6.67 * 10 ^{ -11 }$ $\frac{Nm ^ 2}{kg^2}$