When Emma pushes a bag with a force of 27 newtons, the coefficient of kinetic friction between the bag and the floor is 0.23. Wh

at is the normal force acting on the bag by the floor?
0.085 newtons
27 newtons
2.7 × 102 newtons
1.2 × 102 newtons

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inna [3103]

Answer:

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$ ≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

F(x,y,z) = y i − x j + $z^{2}$ k

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + $v^{2}$ k

The normal vector N is computed as

$N = S_{u}XS_{v}$

Where:

$S_{u} = =$

$S_{v} = =$

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, $v^{2}$ >. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + u $v^{2}$

Thus

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu$ ≈ 3077.34

8 0

2 months ago

Sav [3153]

The element is magnesium (Mg).

7 0

2 months ago