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4 years ago

What is the point of an object where the force of gravity is considered to act?

1 answer:

6 0

Gravity always acts downward on every object on earth. Gravity multiplied by the object's mass produces a force called weight. Although the force of an object's weight acts downward on every particle of the object, it is usually considered to act as a single force through its balance point, or center of gravity.

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Don't get any of this

Serga [27]

Answer:

the value of the printer is going down over the years because it is getting older. all you need to do for this is label the x and y axis with the nbers on the graph and then fill it in

has ur teacher gone over how to fill in graphs yet?

5 0

3 years ago

Read 2 more answers

Charlene wants to build a fence around her rectangular garden. The garden measures 20 meters by 60 meters.

myrzilka [38]

The perimeter because the fence goes AROUND the yard, not in it.

P = 2w + 2l

P = 2(60) + 2(20)

P = 120 + 40

P = 160

Charlene needs 160 meters of fencing material.

3 0

4 years ago

Read 2 more answers

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

4 years ago

The USA won 15 Wimbledon championships. Sweden and Switzerland both won 7. Australia won 6, German won 4, and Spain won 2. Write

mihalych1998 [28]

Well you add the total of championships. 15+7+7+6+4+2=41
The ratio is 6:41
This means that Australia won 6 of the 41 championships.

5 0

3 years ago

An organic farm has been growing an heirloom variety of summer squash. A sample of the weights of 40 summer squash revealed that

natta225 [31]

Answer:

b. 0.0228

Step-by-step explanation:

We are given that

n=40

Mean, $\mu=402.7 g$

Standard deviation, $\sigma=8.8$ g

We have to find the probability hat the mean weight for a sample of 40 summer squash exceeds 405.5 grams.

$P(x>405.5)=P(\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{405.5-402.7}{\frac{8.8}{\sqrt{40}}})$

$P(x>405.5)=P(Z>\frac{2.8}{\frac{8.8}{\sqrt{40}}})$

$P(x>405.5)=P(Z>2.01)$

$P(x>405.5)=1-P(Z\leq 2.01)$

$P(x>405.5)=1-0.977784$

$P(x>405.5)=0.022216$

Hence, option b is correct.

5 0

3 years ago