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2 years ago

(Please Help I will mark Brainliest to the most accurate answer)

Mathematics

2 answers:

harkovskaia [24]2 years ago

8 0

Answer:

8π

Step-by-step explanation:

8π because the circumference is C=dπ

so since d = 8

the circumference is 8π

Hope this helps :3

Alika [10]2 years ago

3 0

Answer:

4 pi cm

Step-by-step explanation:

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Seperate the number 53 into 3 parts so that the second number is 3 less than the first number and the third number is twice the

irga5000 [103]

Hello,

Let a the first, b the second and c the third.
a+b+c=53
b=a/3
c=2a
==>a+a/3+2a=53
==>10a/3=53
==>a=15.9
c=2*a=2*15.9=31.8
b=a/3=15.9/3=5.3

Proof: a+b+c=15.9+5.3+31.8=53

The second number is 5.3

7 0

2 years ago

A star in the sky appears to move by 12 arcseconds over the course of 6 months. Convert this number into first areminutes then d

Bumek [7]

Answer:

Given :A star in the sky appears to move by 12 arc seconds over the course of 6 months.

To Find :Convert this number into first arc minutes then degrees.

Solution:

60 arc seconds = 1 arc minute

1 arc seconds = $\frac{1}{60}$ arc minute

12 arc seconds = $\frac{1}{60} \times 12$ arc minute

12 arc seconds = $\frac{1}{5}$ arc minute

So, 12 arc seconds = $\frac{1}{5}$ arc minute

Now convert it into degrees

60 arc minute = 1 degree

1 arc minute = $\frac{1}{60}$ degree

$\frac{1}{5}$ arc minute = $\frac{1}{60} \times \frac{1}{5}$ degree

$\frac{1}{5}$ arc minute = $\frac{1}{300}$ degrees

12 arc seconds = $\frac{1}{300}$ degrees

8 0

2 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: