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3 years ago

What type of wave is a microwave

Physics

2 answers:

eduard3 years ago

7 0

It would be a electromagnetic wave
Hope that this would help you! =)

AfilCa [17]3 years ago

4 0

Answer: The correct answer is electromagnetic wave.

Explanation:

In the electromagnetic wave, both magnetic field and electric field are perpendiculars each other to the direction of the motion of the wave.

Electromagnetic spectrum is the range of frequencies of electromagnetic radiations.

Microwave is an electromagnetic wave. It falls in the range of electromagnetic spectrum between radio and infrared rays. It is a high bandwidth communication. It is used in RADAR and for heating purpose in microwave oven. Its frequency range lies between 300 MHz and 300 GHz.

Therefore, microwave is the type of electromagnetic wave.

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What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?

Ivahew [28]

The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.

Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima. The latitude in the middle
of that intersection is 46.585° North. That's the number we need.

Here's how I would do it:

-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.

-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°.

-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.

This sets the limit of the highest in the sky that the moon can ever appear.

90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .

That doesn't happen regularly. It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).

Depending on the time of year, that can be any time of the day or night.

The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.

In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky. Then it's going to be somewhere near
67° above the horizon at midnight.

5 0

3 years ago

What does keplers second law of planetary monition imply

Gemiola [76]

Kepler's second law of planetary motion describes the speed of a planet traveling in an elliptical orbit around the sun. It states that a line between the sun and the planetsweeps equal areas in equal times. Thus, the speed of theplanet increases as it nears the sun and decreases as it recedes from the sun.

6 0

3 years ago

Read 2 more answers

Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25

mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp = 1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0

3 years ago

Help please, I need it

love history [14]

Answer:

1.97×10⁻²¹ J

Explanation:

Use ideal gas law to find temperature.

PV = nRT

(9 atm) (9 L) = (83.3 mol) (0.0821 L·atm/mol/K) T

T = 11.9 K

The average kinetic energy per atom is:

KE = 3/2 kT

KE = 3/2 (1.38×10⁻²³ J/K) (11.9 K)

KE = 2.46×10⁻²² J

For a mass of 5.34×10⁻²⁶ kg, the kinetic energy is:

KE = (5.34×10⁻²⁶ kg) (1 mol / 0.004 kg) (6.02×10²³ atom/mol) (2.46×10⁻²² J)

KE = 1.97×10⁻²¹ J

5 0

4 years ago

An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a

Morgarella [4.7K]

Answer:

0.37 m

Explanation:

The angular frequency, ω, of a loaded spring is related to the period, T, by

$\omega = \dfrac{2\pi}{T}$

The maximum velocity of the oscillation occurs at the equilibrium point and is given by

$v = \omega A$

A is the amplitude or maximum displacement from the equilibrium.

$v = \dfrac{2\pi A}{T}$

From the the question, T = 0.58 and A = 25 cm = 0.25 m. Taking π as 3.142,

$v = \dfrac{2\times3.142\times0.25\text{ m}}{0.58\text{ s}} = 2.71 \text{ m/s}$

To determine the height we reached, we consider the beginning of the vertical motion as the equilibrium point with velocity, v. Since it is against gravity, acceleration of gravity is negative. At maximum height, the final velocity is 0 m/s. We use the equation

$v_f^2 = v_i^2+2ah$

$v_f$ is the final velocity, $v_i$ is the initial velocity (same as v above), a is acceleration of gravity and h is the height.

$h = \dfrac{v_f^2 - v_i^2}{2a}$

$h = \dfrac{0^2 - 2.71^2}{2\times-9.81} = 0.37 \text{ m}$

3 0

3 years ago