How are physical and chemical changes similar?

Why are infrared waves located between microwaves and visible light on the electromagnetic spectrum?

dangina [55]

Answer: The infra red waves is located between microwave and visible light based on their WAVELENGTH and FREQUENCY of occurrence.

Explanation:

Electromagnetic waves are those waves that do not require or need a material medium for its propagation, but they are able to travel through a vacuum. They exhibit or show all properties associated or connected with light. They are undeflected in electric and magnetic fields. These electromagnetic waves are arranged in order of their FREQUENCY and WAVELENGTHS which is known as ELECTROMAGNETIC SPECTRUM.

FREQUENCY is defined as the number of cycles which the wave completes in one second and is measured in Hertz(Hz). While WAVELENGTH is defined as the distance between two successive crests or troughs of waves which is measured in meter (m).

The electromagnetic spectrum is made up of the following rays which is arranged from the biggest wavelengths to the smallest:

--> Radiowaves

--> microwave :

--> infrared rays:

--> visible light:

--> ultraviolet rays

--> x-rays and

--> Gamma rays.

According to the arrangement of the spectrum above, the microwave has a higher wavelength and frequency than the infrared rays, while the visible light has a lower wavelength and frequency than the infrared rays.

3 0

2 years ago

Second<br>class lever short note

seraphim [82]

Answer:

wow

Explanation:

6 0

3 years ago

Read 2 more answers

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

puteri [66]

Answer:

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is $0\ mph\ to\ 70\ mph$ .

Initial velocity of the car, $v_i=0\ mph$

final velocity of the car during the test, $v_f=32\ mph=14.3052\ m.s^{-1}$

Time taken to accelerate form zero to 32 mph at full power, $t=1.2\ s$

initial velocity of the car, $u_i=0\ mph$

final desired velocity of the car, $u_f=64\ mph=28.6105\ m.s^{-1}$

Now the acceleration of the car:

$a=\frac{v_f-v_i}{t}$

$a=\frac{14.3052-0}{1.2}$

$a=11.921\ m.s^{-1}$

Now using the equation of motion:

$u_f=u_i+a.t$

$64=0+11.921\times t$

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0

3 years ago

Two identical satellites are in orbit about the earth. One orbit has a radius r and the other 2r. The centripetal force on the s

velikii [3]

Answer:

the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

Explanation:

Mass of satellite, m

orbit radius of first, r1 = r

orbit radius of second, r2 = 2r

Centripetal force is given by

$F= \frac{mv^{2}}{r}$

Where v be the orbital velocity, which is given by

$v=\sqrt{gr}$

So, the centripetal force is given by

$F= \frac{mgr}}{r}}=mg$

where, g bet the acceleration due to gravity

$g=\frac{GM}{r^{2}}$

So, the centripetal force

$F= \frac{GMm}}{r^{2}}}$

Gravitational force on the satellite having larger orbit

$F= \frac{GMm}{4r^{2}}$ .... (1)

Gravitational force on the satellite having smaller orbit

$F'= \frac{GMm}{r^{2}}$ .... (2)

Comparing (1) and (2),

F' = 4 F

So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

8 0

3 years ago

Write the equation expressing the relationship "y varies directly as x." use k as the constant of proportionality.

Alexeev081 [22]

The equation expressing the statement "y varies directly as x" is $y=kx.$

Explanation

The statement that $y$ varies directly as $x$ is analogous to saying that the ratio of $y$ to $x$ is constant. In other words, when $x$ increases, $y$ likewise increases and that when $x$ decreases, $y$ decreases proportionally.

Mathematically, we express the relationship that the ration of $y$ is to $x$ is constant is expressed as; $\frac{y}{x} =k$ where $k$ is the constant of proportionality.

We can then solve the relationship for $y$ to determine the correct form of the relationship as shown below,

$\frac{y}{x} =k\\\rightarrow y=kx$

6 0

3 years ago

Read 2 more answers