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2 years ago

If a student applied force 36N on a bench and it displaced by 4 m how much work did that student do

Mathematics

2 answers:

MAVERICK [17]2 years ago

8 0

<h3>Answer: 144 Joules </h3>

Steps Shown:

Work = Force * Displacement

Work = (36 Newtons)*(4 meters)

Work = (36*4) Newton meters

Work = 144 Newton meters

Work = 144 Joules

Note: The displacement must point in the same direction as the force applied.

castortr0y [4]2 years ago

5 0

Answer:

The answer is 144

Step-by-step explanation:

Work Done = Force × Distance

W= F×D

F= 36N

D= 4m

W= 36×4

W= 144j

N/B The workdone is always in Joules.

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3 years ago

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Viefleur [7K]

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3 years ago

Parallelogram ABCD has the angle measures shown. Can you conclude that it is a rhombus, a rectangle, or a square? Explain.

Hoochie [10]

Answer:

The correct option is parallelogram ABCD is a rhombus, because the diagonal bisects two angles

Step-by-step explanation:

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∠B = ∠D

Thus AB=AC by the property of opposite sides of equal angles are equal

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∠B = ∠D

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3 years ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial: