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Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9) The amount of work required is ____ (Simplify your answer.)

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

We want to compute the curve integral (or line integral)

$\bf \int_(C)F$

where the force field F is defined by

F(x,y) = (y, -x)

and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9).

We can write

C = $\bf C_1+C_2$

where

$\bf C_1$ = line segment from (1, 5) to (0, 0)

$\bf C_2$ = line segment from (0, 0) to (0, 9)

so,

$\bf \int_(C)F=\int_(C_1)F+\int_(C_2)F$

Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with

r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1

Hence $\bf C_1$ can be parameterized as

$\bf r_1(t) = (1-t, 5-5t)$ for 0 ≤ t ≤ 1

and $\bf C_2$ can be parameterized as

$\bf r_2(t) = (0, 9t)$ for 0 ≤ t ≤ 1

The derivatives are

$\bf r_1'(t) = (-1, -5)$

$\bf r_2'(t) = (0, 9)$

and

$\bf \int_(C_1)F=\int_(0)^(1)F(r_1(t))\circ r_1'(t)dt=\int_(0)^(1)(5-5t,t-1)\circ (-1,-5)dt=0$

$\bf \int_(C_2)F=\int_(0)^(1)F(r_2(t))\circ r_2'(t)dt=\int_(0)^(1)(9t,0)\circ (0,-9)dt=0$

In consequence,

$\bf \int_(C)F=0$

Answer 2

Answer:

Final answer:

The work done by the force field F = (y, -x) along the given path is -5 Joules. This was calculated by breaking the path into two segments and calculating the work done for each segment.

Explanation:

To calculate the work done by the force field F = (y, -x) when moving an object along a specific path, we utilize the concept of the line integral or the dot product of the force and the displacement vector. We can break down the given path into two line segments and solve each separately.

The first segment is from (1, 5) to (0, 0). Only the x component of the displacement is non-zero here, the force is F = (5, -1). Thus the work done on this segment is given by W = F.d = Fd cos θ = -(5 N)(1 m)(cos(180)) = -5 J, where J stands for Joules, the unit of work or energy.

The second segment is from (0, 0) to (0, 9). The force and displacement are perpendicular so the work done is 0.

By adding the work done on these two segments, we arrive at the total work done: W_total = -5 J + 0 J = -5 J

Learn more about Work done by a force field here:

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Answers

Answer:

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Step-by-step explanation:

Given rectangle JKLM

As per graph, the rectangle J'K'L'M' is the transformation of JKLM

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Trapezoid STUV:

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Answers

Answer:

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Step-by-step explanation:

From the given information;

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Experiment a), where you need four 3s is a Bernoulli trial, as getting a 3 is sucess and not getting a 3 is a failure, and each roll of the dice is independent from each other.

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