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Allisa
1 month ago
11

A 45.0-kg person steps on a scale in an elevator. The scale reads 460 n. What is the magnitude of the acceleration of the elevat

or?
Physics
1 answer:
ValentinkaMS [3.4K]1 month ago
4 0
The elevator's acceleration is 0.422 m/s². To clarify the solution: By applying Newton's Law, the net forces in the motion's direction equal the mass multiplied by the acceleration. The forces comprise 460 N in the motion's direction and the person's weight acting in the opposite direction... The weight is determined by the mass and gravity's acceleration (W = mg). Here m = 45 kg and g = 9.8 m/s², leading to W = 441 N. With the scale indicating 460 N, we apply F - W = ma, yielding 19 = 45 a. Dividing both sides by 45 gives a = 0.422 m/s².
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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
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 + uv^{2}

Thus

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

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