You are sitting on the beach and wondering about the properties of mechanical waves. Describe them in terms of ocean waves.

Physics

1 answer:

Rudiy273 years ago

7 0

Answer:

The ocean waves is a mechanical wave that transmits mechanical energy in the wave by the synchronized and repeated oscillation of the waters about an equilibrium level such that as the wave approaches the shoreline, and the water depth decreases, the height of the wave also increases reflecting the effective transmission of energy while the medium which is the water through which the wave propagates, move back and forth within a small region

Explanation:

A mechanical wave like other waves is the oscillation of a field about an equilibrium level. In mechanical waves, the field consists of the oscillating matter such that the wave transmits energy through a medium. The displacement of the medium through which the wave energy is limited such that the wave energy is conserved to travel far.

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A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$