TESE

Waves that move the particles of the medium parallel to the direction in which the waves are traveling are called

Nikolay [14]

1. a. longitudinal waves.

There are two types of waves:

- Transverse waves: in transverse waves, the oscillations of the wave occur in a direction perpendicular to the direction of propagation of the wave

- Longitudinal waves: in longitudinal waves, the oscillations of the waves occur parallel to the direction in which the waves are travelling.

So, these types of waves are called longitudinal waves.

2. d. a medium

There are two types of waves:

- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field, and they can travel both in a medium and also in a vacuum (they do not need a medium to propagate)

- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, so they need a medium to propagate - therefore, the correct choice is d. a medium

3. a. AM/FM radio

Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.

5 0

2 years ago

Which transition metals have more in common than other

jeka94

Answer:

A. alkali metals

Explanation:

im not sure about that

8 0

3 years ago

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A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A

Jet001 [13]

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

$F = ILBsin \theta$

Here,

I = Current

L = Length

B = Magnetic Field

$\theta$ = Angle between Force and Magnetic Field

But $\theta = 90\°$

$F = ILB$

Rearranging to find the Magnetic Field,

$B = \frac{F}{IL}$

Here the force per unit length,

$B = \frac{1}{I}\frac{F}{L}$

Replacing with our values,

$B = \frac{0.130N/m}{16.5}$

$B = 0.0078T$

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

6 0

3 years ago

A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit

djverab [1.8K]

Answer:

The current is $I = 6.68 \ A$

Explanation:

From the question we are told that

The radius of the loop is $r = 6 \ cm = 0.06 \ m$

The earth's magnetic field is $B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T$

The number of turns is $N =1$

Generally the magnetic field generated by the current in the loop is mathematically represented as

$B = \frac{\mu_o * N * I}{2 r }$

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

$B = B_e$

=> $B_e = \frac{\mu_o * N * I }{ 2 * r}$

Where $\mu$ is the permeability of free space with value $\mu _o = 4\pi * 10^{-7} N/A^2$

$0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}$

=> $I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}$

$I = 6.68 \ A$

3 0

3 years ago

Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6

Nana76 [90]

Answer:

The answer to the question is as follows

The acceleration due to gravity for low for orbit is 9.231 m/s²

Explanation:

The gravitational force is given as

$F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }$

Where $F_{G}$ = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹ $\frac{Nm^{2} }{kg^{2} }$

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by $F_{G}$ and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and $F_{G}$ = $\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }$ = 9.231 N

Therefore the acceleration due to gravity = $F_{G}$ /mass

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²

4 0

3 years ago

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