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2 years ago

Waves begin to "feel bottom" when the depth of water is

2 answers:

8 0

I think the correct answer would be one half the wavelength. Waves would "feel bottom" when the water is at the depth of 0.5 of the wavelength. "Feel bottom" is a term used to describe that the depth of water affects the wave properties. Hope this answers the question.

Likurg_2 [28]2 years ago

6 0

Equal to one-half the wavelength is the right answer.

Water is regulated by many approaches such as depth gauges (based on force) a sounding line and by sonar range. Prior to the modern era, sailors would cast metal weighted materials into the ocean. Submarines made a new demand for depth estimation, to need to know the depth of boats in the water. However, waves start to "feel bottom" when the depth of water is equivalent to one-half the wavelength. As the wave approaches a lightweight bank, its track is adjusted. This involution made the tides to become inactive. As the flow becomes steeper, the width of the wave diminishes.

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An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

viktelen [127]

Answer:

$1.232\times 10^{-5}\ T.$

Explanation:

<u>Given:</u>

Current through the wire, passing through the origin, $I_1 = 250\ A.$

Current through the wire, passing through the y axis, $r_y=1.8\ m.$ , $I_2 = 50\ A.$

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to $\mu_o$ times the net current threading the loop.

$\oint \vec B \cdot d\vec l=\mu_o I.$

In case of a circular loop, the directions of magnetic field and the line element $d\vec l$ , both are along the tangent of the loop at that point, therefore, $\vec B\cdot d\vec l = B\ dl$ .

$\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\$

$\oint dl$ is the circumference of the Amperian loop = $2\pi r$

Therefore,

$B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.$

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_1 = 3.510\ m.$

The magnetic field at the given point due to this wire is given by:

$B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.$

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_2 = 5.310\ m.$

The magnetic field at the given point due to this wire is given by:

$B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.$

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at $r_r=-3.510\ m$ is given by

$B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.$

6 0

3 years ago

A swimming duck paddles the water with its feet once every 1.6 s, producing surface waves with this period. The duck is moving a

emmainna [20.7K]

Answer:

0.245 m/s

0.904 m

Explanation:

$v_{d}$ = speed of duck ahead of wave

$v_{s}$ = speed of surface wave = 0.32 m/s

T = time for paddling = 1.6 s

d = spacing between the waves = 0.12 m

speed of duck ahead of wave is given as

$v_{d}$ = $v_{s}$ - $\frac{d}{T}$

$v_{d}$ = 0.32 - $\frac{0.12}{1.6}$

$v_{d}$ = 0.245 m/s

$v_{w}$ = speed of wave behind the duck

speed of wave behind the duck is given as

$v_{w}$ = $v_{s}$ + $v_{d}$

$v_{w}$ = 0.32 + 0.245

$v_{w}$ = 0.565 m/s

D = spacing between the crests

spacing between the crests is given as

D = $v_{w}$ T

D = (0.565) (1.6)

D = 0.904 m

7 0

2 years ago

Which particle of the atom has no charge? A. proton B. electron C. nucleus D. neutron

Fynjy0 [20]

D
Neutron is <span>a subatomic particle of about the same mass as a proton but without an electric charge, present in all atomic nuclei except those of ordinary hydrogen.</span>

8 0

2 years ago

When an object is put into a liquid, it experiences a buoyant force that is equal to the weight of the liquid the object displac

Anon25 [30]

Answer:

A) F = 1.8375 N, B) V = 1.25 10⁻⁴ m³, C) V= 1.25 10⁻⁴ m³

Explanation:

This exercise should select some values such as the weight of the block m=0.250 kg and the density of the liquid ρ = 500 kg / m³ (water); with these values answer the question, for this we use the static equilibrium relations.

Σ F = 0

F + B - W = 0

the expression for thrust is

B = ρ g V_liquid

A) the weight of the block is

W = m g

W = 0.250 9.8

W = 2.45 N

Let's look for the maximum push

V_body = V_liquid

B = 500 9.8 0.05³

B = 0.6125 N

we substitute

F = W - B

F = 2.45 - 0.6125

F = 1.8375 N

B) As the body is totally submerged, the volume of the liquid and the body are equal

V = l³

V = 0.05³

V = 1.25 10⁻⁴ m³

C) the volume of water displaced is equal to the volume of the body

V_liquid = 1.25 10⁻⁴ m³

7 0

2 years ago

What do the single electrons in nitrogen do for a triple bond?

kirza4 [7]

Nitrogen could form 3 bonds based on octet rule, because it has 5 valence electrons. That means it needs 3 bonds.

Explanation:

A nitrogen atom can fill its octet by sharing three electrons with another nitrogen atom, forming three covalent bonds, a so-called triple bond.
A triple bond isn't quite three times as strong as a single bond, but it's a very strong bond.
Every covalent bond is a sharing of two electrons between two atoms. A double bond is 4 electrons being shared (2x2). Therefore a triple bond is 6 electrons being shared (2x3)
Triple bonds are stronger than double bonds due to the the presence of two pi bonds rather than one.
Each carbon has two sp hybrid orbitals, and one of them overlaps with its corresponding one from the other carbon atom to form an sp-sp sigma bond.
A single lone pair can be found with atoms in the nitrogen such as nitrogen in ammonia , two lone pairs can be found with atoms in the chalogen group such as oxygen in water and the halogen can carry three lone pairs such as in hydrogen chloride. Nitrogen has 2 lone pairs.

6 0

3 years ago