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Anna71 [15]
3 years ago
12

An electric dipole is made of two charges of equal magnitudes and opposite signs. The positive charge, q=1.0 μC, is located at t

he point (x, y, z) = (0.00 cm, 1.0 cm, 0.00 cm), while the negative charge is located at the point (x, y, z) = (0.00 cm, −1.0 cm, 0.00 cm). How much work will be done by an electric field E⃗ =3.0×106 N/C) i^ to bring the dipole to its stable equilibrium position?
Engineering
1 answer:
vfiekz [6]3 years ago
8 0

Answer:

work done by electric field  is 0.06 J

Explanation:

Given data:

Two point charge is + 1\mu C  and -1 \mu C

0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)

-1 charge positioned is (0 cm , -1 cm, 0.00 cm)

E = 3.0\times 10^6 N/C

From above information, the distance between  given two charges d = 2 cm

then d = 0.02m

 work needed is W = q E d

W = 1.0 \times 10^{-6} \times 3.0 \times 10^6 \times 0.02

W = 0.06 J  

Therefore work done by electric field  is 0.06 J

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Prove the following languages are nonregular, once using the pumping lemma and once using the Myhill-Nerode theorem. When using
VashaNatasha [74]

Answer:

For any string, we use s = xyz

Explanation:

The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.

Here are the cases:

  • Consider any string a i b j c k in the language. If i = 1 or i > 2, we take x = \epsilon   and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.
  • For i = 2, we can take    and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
  • Finally, for the case i = 0, we take x = \epsilon  , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.
8 0
3 years ago
A rotor in a compressor stage has a mean blade radius of 0.285 m and an angular rotor velocity of 8500 RPMs. The static temperat
pantera1 [17]

Answer:

0,285 is the answer

Explanation:

6 0
3 years ago
The waffle slab is: a) the two-way concrete joist framing system. b) a one-way floor and roof framing system. c) the one-way con
GREYUIT [131]

Answer:

a) the two-way concrete joist framing system

Explanation:

A waffle slab is also known as ribbed slab, it is a slab which as waffle like appearance with holes beneath. It is adopted in construction projects that has long length, length more than 12m. The waffle slab is rigid, therefore it is used in building that needs minimal vibration.

4 0
3 years ago
A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b
Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions  = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition  Cβ = 87  

we use here lever rule that is

Wα = Wβ   ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

\frac{C_\beta - Co}{C_\beta - Ca}   = 0.5

\frac{87 - 57}{87 - Ca} = 0.5  

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0
3 years ago
The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm
aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

T_max =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=14.5MPa

T_{min} =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=10.3MPa

7 0
3 years ago
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