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3 years ago

15

A chain 18 feet long whose weight is 84 pounds is hanging over the edge of a tall building and does not touch the ground. How mu

ch work is required to lift the entire chain to the top of the building

1 answer:

coldgirl [10]3 years ago

8 0

Answer:

756 feet pound

Step-by-step explanation:

Length of the chain, L = 18 feet

weight of chain, W = 84 pound

If we lift the centre of gravity of the chain, it will entirely lift up. The centre of gravity of the uniform chain is at its mid point. So, the work done

W = weight x half of length

W = 84 x 9

W = 756 feet pound

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Find the average rate of change of g(x) = 1x3 + 2 from x =<br> 1 to x = 4

Oksi-84 [34.3K]

Answer:

Step-by-step explanation:

Explanation:

The

average rate of change

of g(x) over an interval between 2 points (a ,g(a)) and (b ,g(b) is the slope of the

secant line

connecting the 2 points.

To calculate the average rate of change between the 2 points use.

∣

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

g

(

b

)

−

g

(

a

)

b

−

a

a

a

∣

∣

∣

−−−−−−−−−−−−−−−

g

(

6

)

=

6

2

−

6

+

3

=

33

and

g

(

4

)

=

4

2

−

4

+

3

=

15

Thus the average rate of change between (4 ,15) and (6 ,33) is

33

−

15

6

−

4

=

18

2

=

9

This means that the average of all the slopes of lines tangent to the graph of g(x) between (4 ,15) and (6 ,33) is 9

3 0

2 years ago

The coordinates of the endpoints of ST are (-2,3) and (3, y) respectively supposethe distance between s and t us 13 units.What v

igor_vitrenko [27]

Answer:

The values of y would be -9 and 15

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$d=13\ units$

S(-2,3) and T(3,y)

substitute the given values in the formula and solve for y

$13=\sqrt{(y-3)^{2}+(3+2)^{2}}$

$13=\sqrt{(y-3)^{2}+25}$

squared both sides

$169=(y-3)^{2}+25$

$(y-3)^{2}=169-25$

$(y-3)^{2}=144$

take square root both sides

$y-3=(+/-)12$

$y=3(+/-)12$

$y=3(+)12=15$

$y=3(-)12=-9$

therefore

The values of y would be -9 and 15

5 0

3 years ago

Round 7,883 to nearest thousand

Stels [109]

Answer:

8,000

Step-by-step explanation:

the nearest thousand is 8,000 as it is past 7,500.

6 0

3 years ago

Read 2 more answers

An engineer deposits $2,000 per month for four years at a rate of 24% per year, compounded semi-annually. How much will he be ab

Alex17521 [72]

The amount of money he will be able to withdraw after 10 years after his last deposit is $926,400.

<h3>Compound interest</h3>

Principal, P = $2,000 × 12 × 4

= $96,000

Time, t = 10 years
Interest rate, r = 24% = 0.24
Number of periods, n = 2

A = P(1 + r/n)^nt

= $96,000( 1 + 0.24/2)^(2×10)

= 96,000 (1 + 0.12)^20

= 96,000(1.12)^20

= 96,000(9.65)

= $926,400

Therefore, the amount of money he will be able to withdraw after 10 years after his last deposit is $926,400

Learn more about compound interest:

brainly.com/question/24924853

#SPJ4

5 0

1 year 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:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

8 0

2 years ago