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nlexa [21]
3 years ago
13

What do radio waves and microwaves have in common?

Physics
2 answers:
barxatty [35]3 years ago
7 0

Answer:

Both are found next to visible light on the electromagnetic spectrum.

i know because i just took the test trust me i choose Both are found next to visible light on the electromagnetic spectrum. and got it wrong it said the answer was Both are found next to visible light on the electromagnetic spectrum.

Explanation:

forsale [732]3 years ago
3 0

Answer:

Both are at the side of the spectrum that has the lowest frequency.

Explanation:

We know that the electromagnetic spectrum is the set of several waves that are arranged in order of increasing wavelength of decreasing frequency. It consists of many waves like,

gamma rays, x- rays, ultraviolet rays, visible, infrared rays, microwave, radio waves.

Radio wave has the longest wavelength while gamma rays has maximum wavelength. From the attached figure, it is clear that the frequency of radio wave is least and that of microwave is second least frequency.

Hence, the correct option is (c).

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Please help me out i'm so depressed and such a failure
Anika [276]

Answer: a variety of ohmic valu example, VIN = VR1 + VR2.

Potentiometer Example No1

A resistor of 250 ohms is connected in series with a second resistor of 750 ohms so that the 250 ohm resistor is connected to a supply of 12 volts and the 750 ohm resistor is connected to ground (0v). Calculate the total series resistance, the current flowing through the series circuit and the voltage drop across the 750 ohm resistor.

potentiometer example one

 

Explanation:

uman ear has a logarithmic response and is therefore non-linear.

If we where to use a linear potentiometer to control the volume, it would give the impression to the ear that most of the volume adjustment was restricted to one end of the pots track. The logarithmic potentiometer however, gives the impression of a more even and balanced volume adjustment across the full rotation of the volume control.

So the operation of a logarithmic potentiometers when adjusted is to produce an output signal which closely matches the nonlinear sensitivity of the human ear making the volume level sound as though it is increasing linearly. However, some cheaper logarithmic potentiometers are more exponential in resistance changes rather than logarithmic but are still called logarithmic because their resistive response is linear on a log scale. As well as logarithmic potentiometers, there are also anti-logarithmic potentiometers in which their resistance quickly increases initially but then levels off.

The all potentiometers and rheostats are available in a choice of different resistive tracks or patterns, known as laws, being either linear, logarithmic, or anti-logarithmic. These terms are more commonly abbreviated to lin, log, and anti-log, respectively.

The best way to determine the type, or law of a particular potentiometer is to set the pots shaft to the center of its travel, that is about half way, and then measure the resistance across each half from wiper to end terminal. If each half has more or less equal resistance, then it’s a Linear Potentiometer. If the resistance appears to be split at about 90% one way and 10% the other then chances are it’s a Logarithmic Potentiometer.

Potentiometer Summary

In this tutorial about potentiometers, we have seen that a potentiometer or variable resistor basically consists of a resistive track with a connection at either end and a third terminal called the wiper with the position of the wiper dividing the resistive track. The position of the wiper on the track is adjusted mechanically by rotating a shaft or by using a screwdriver.

Variable resistors can be categorised into one of two operational modes – the variable voltage divider or the variable current rheostat. The potentiometer is a three terminal device used for voltage control, while the rheostat is a two terminal device used for current control.

We can summarise this in the following table:

Type Potentiometer Rheostat

Number of

Connections Three Terminals Two Terminals

Number of Turns Single and Multi-turn Single-turn Only

Connection Type Connected Parallel with a Voltage Source Connected in Series with the Load

Quantity Controlled Controls Voltage Controls Current

Type of Taper Law Linear and Logarithmic Linear Only

Then the potentiometer, trimmer and rheostat are electromechanical devices designed so that their resistance values can be easily changed. They can be designed as single-turn pots, presets, slider pots, or as multi-turn trimmers. Wirewound rheostats are mainly used to control an electrical current. Potentiometers and rheostats are also available as multi-gang devices and can be classified as having either a linear taper or a logarithmic taper.

Either way, potentiometers can provide highly precise sensing and measurement for linear or rotary movement as their output voltage is proportional to the wipers position. The advantages of potentiometers include low cost, simple operation, lots of shapes, sizes and designs and can be used in a vast array of different applications.

However as mechanical devices, their disadvantages include eventual wear-out of the sliding contact wiper and/or track, limited current handling capabilities (unlike Rheostats), electrical power restrictions and rotational angles that are limited to less than 270 degrees for single turn pots

5 0
3 years ago
Una moto accelera da ferma a 80 km/h in 4,0 s. A quale forza di inerzia è sottoposto il pilota che ha massa 75 kg?
scoray [572]

A=65 divideded by 55 plus 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Explanation:

4 0
3 years ago
Two stones are dropped from the edge of a 60m cliff , the second stone 1.6secon after the first . How far below the top of the c
tigry1 [53]

Answer:

The separation between the two stones is 36 m, when the second stone is approximately 10.9 m below the top of the cliff

Explanation:

The given parameters are;

The height of the cliff from which the stones are dropped, h = 60 m

The time at which the second stone is dropped = 1.6 seconds after the first

The distance below the top of the cliff when the distance between the two stones is 36 m = Required

We have;

The kinematic equation of motion that can be used is s = u·t - (1/2)·g·t²

For the first stone, we have, s₁ = u·t₁ - (1/2)·g·t₁²

For the second stone, we get; s₂ = u·t₂ - (1/2)·g·t₂²

t₁ = t₂ + 1.6

g = The acceleration due to gravity ≈ 9.81 m/s²

s = The distance below the cliff top

The initial velocity of the stones, u = 0

Let<em> t</em> represent the time from which the second stone is dropped at which the distance between the two stones is 36 m, we have;

s₁ = u·(t + 1.6) + (1/2)·g·(t + 1.6)²

s₂ = u·t + (1/2)·g·t²

u = 0

∴ s₁ - s₂ = 36 =  (1/2)·g·(t + 1.6)² - (1/2)·g·t²

2 × 36/(g) = (t + 1.6)² - t²  = t² + 3.2·t + 2.56 - t² = 3.2·t + 2.56

2 × 36/(9.81) = 3.2·t + 2.56

t = (2 × 36/(9.81) - 2.56)/3.2 =  ≈ 1.49 s

t ≈ 1.49 s

s₂ = (1/2)·g·t²

∴ s₂ = (1/2) × 9.81 × 1.49² ≈ 10.9

The distance below the top of the cliff of the second stone when the the separation between the two stones is 36 m, s₂ ≈ 10.9 m.

5 0
3 years ago
If a bar magnet's neutral region is broken into two, what will most likely occur?
sattari [20]
You can think of a magnet<span> as a bundle of tiny </span>magnets<span>, </span><span> that are jammed together. Each one reinforces the </span>magnetic<span> fields of the others. Each one has a tiny north and south pole. </span>If you cut<span> one in </span>half<span>, the newly </span>cut<span> faces will become the new north or south poles of the smaller pieces.
</span>
So the answer is D

4 0
4 years ago
Read 2 more answers
A bullet of mass 20g is horizontally fired with a velocity of 150m/s from a pistol of mass 2 kg. What is the
kiruha [24]

Answer:

m1 = 20g (= 0.02 kg)

Mass of pistol, m2 = 2 kg

Initial velocity of the bullet (u1) and pistol (u2) = 0

Final velocity of the bullet, v1 = +150m s-1

Let v be the recoil velocity of the pistol.

Total momentum of the pistol and bullet after it is fired is

= (0.02 kg x 150 m s-1) + (2 kg x v m s-1)

= (3 + 2v) kg m s-1

Total momentum after the fire = Total momentum before the fire

3 + 2v = 0

→v = -1.5 m/s  

7 0
3 years ago
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