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Zarrin [17]
3 years ago
13

Given that curl F = 2yi – 2zj + 3k, find the surface integral of the normal component of curl F (not F) over (a) the open hemisp

herical surface x2 + y2 + z2 = 9, z > 0. (b) the sphere x2 + y2 + z2 = 9. (In both parts, you should be able to write the answer down by inspection.)
Mathematics
1 answer:
Dimas [21]3 years ago
7 0

Use Stokes' theorem for both parts, which equates the surface integral of the curl to the line integral along the surface's boundary.

a. The boundary of the hemisphere is the circle x^2+y^2=9 in the plane z=0, where the curl is \mathrm{curl}\vec F=2y\,\vec\imath+3\,\vec k. Green's theorem applies here, so that

\displaystyle\iint_S\mathrm{curl}\vec F\cdot\mathrm d\vec S=\int_{\partial S}\vec F\cdot\mathrm d\vec r=3\int_{x^2+y^2=9}\mathrm d\vec r

which means the value of the line integral is 3 times the area of the circle, or 27\pi.

b. The closed sphere has no boundary, so by Stokes' theorem the integral is 0.

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Find the mean absolute deviation of the set of data 3,8,10,12,15
inn [45]

Answer:

3.28

Step-by-step explanation:

find the mean of all the numbers

3 + 8 + 10 + 12 + 15 = 48

48/5=9.6

so the mean of all the numbers is 9.6

now subtract 9.6 from all the numbers

|3 - 9.6| = 6.6

|8 - 9.6| = 1.6

|10 - 9.6| = .4

|12 - 9.6| = 2.4

|15 - 9.6| = 5.4

Find the mean/average of the new values

6.6 + 1.6 + .4 + 2.4 + 5.4 = 16.4

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3 years ago
A cylinder is sliced by a plane that is not parallel to the base
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4 0
3 years ago
Suppose a student started with 139.0 mg of trans-cinnamic acid, 463 mg of pyridinium tribromide, and 2.45 mL of glacial acetic a
Sliva [168]

Answer:

0.2889 g  brominated product

64.6 %

Step-by-step explanation:

This is a bromination chemical reaction of an alkene  and we are asked to calculate the theoretical and percent yield. Thus to solve it we have to perform a calculation based on the balanced chemical reaction.

The pyridinium tribromide is used as a generator of molecular bromine in situ, and bromine will add to the double bond in trans-cinnamic acid so we know the reaction occur in a one to one mole fashion.

trans-cinnamic acid + pyridinium  tribromide ⇒ 2,3-dibromo-3-                          

                                                                             phenylpropanoic acid

Molar weight  trans-cinnamic acid  =  148.16 g/mol

mass trans-cinnamic acid  = 139.0 mg x  1g/1000 mg = 0.139 g

# mol trans-cinnamic acid =  0.139 g / 148 g/mol = 9.38 x 10⁻⁴ mol

Since our reaction is 1 mol trans-cinnamic acid  produces 1 mol 2,3-dibromo-3-phenylpropanoic acid, it follows that the theoretical yield is:

1 mol 2,3-dibromo-3-phenylpropanoic acid /  trans-cinnamic acid x 9.38 x 10⁻⁴ mol trans-cinnamic acid  

=  9.38 x 10⁻⁴ mol  2,3-dibromo-3-phenylpropanoic acid

In grams the the theoretical yield is:

molar mass  2,3-dibromo-3-phenylpropanoic acid = 307.97 g/mol

The theoretical mass  2,3-dibromo-3-phenylpropanoic acid:

=   9.38 x 10⁻⁴ mol  2,3-dibromo-3-phenylpropanoic acid x 307.97 g/mol

=   0.2889 g

% yield = mass experimental/mass theoretical

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4 0
3 years ago
Please help with the second question thank you
boyakko [2]

Answer:

The correct answer is C.

Step-by-step explanation:

In order to solve this problem, you need to simplify the equation by using a natural log (ln). This will cancel out the e and make it easy to solve.

e^(2x + 5) = 4 ----> Take the ln

lne^(2x + 5) = ln(4) ------> Simplify

2x + 5 = ln(4) ------> Subtract 5

2x = ln(4) - 5 -----> Divide by 2

x = [ln(4) - 5]/2

5 0
3 years ago
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