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S_A_V [24]
3 years ago
10

A surface wave is a combination of what two waves?

Physics
2 answers:
Elena-2011 [213]3 years ago
8 0

A for sure, just took the quiz:)

kotegsom [21]3 years ago
4 0

transverse and longitudinal

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An electric dipole consisting of charges of magnitude 2.00 nC separated by 8.40 μm is in an electric field of strength 1390 N/C.
amid [387]

Answer:

(a) The magnitude of the electric dipole moment is 1.68 x 10⁻¹⁴ C.m

(b) The difference between the potential energies ΔU, is 4.6704 x 10⁻¹¹ J

Explanation:

Given;

magnitude of charge, q = 2 nC = 2 x 10⁻⁹ C

distance of separation, d = 8.4 μm = 8.4 x 10⁻⁶ m

strength of electric field, E = 1390 N/C

(a) the magnitude of the electric dipole moment

p = qd

p = (2 x 10⁻⁹ C)(8.4 x 10⁻⁶ m)

p = 1.68 x 10⁻¹⁴ C.m

(b) the difference between the potential energies for dipole orientations parallel and anti-parallel to E

ΔU = U(180) - U(0)

ΔU = 2pE

ΔU = 2(1.68 x 10⁻¹⁴ )(1390)

ΔU = 4.6704 x 10⁻¹¹ J

6 0
3 years ago
If an aircraft is flying at 400 miles/hour and encounters a head wind of 40 miles/hour, what is the resultant ground speed?
Katen [24]
Look at it this way:

Every time the plane flies through 400 miles of air,
the air moves it 40 miles backwards.
So it only covers 360 miles of ground in an hour.
8 0
3 years ago
A hydrogen atom has its electron in the n = 6 level. the radius of the electron's orbit in the bohr model is 1.905 nm.
gayaneshka [121]
Are there any options or is it not multiple choice.
5 0
3 years ago
Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?
nydimaria [60]

Answer:

15~miles

Explanation:

2\frac{1}{2}*6=\frac{2*2+1}{2}*6=\frac{5}{2}*6=\frac{30}{2}=15~miles

7 0
3 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
3 years ago
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