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3 years ago

6

Which of the following types of electromagnetic radiation has the shortest frequency? (2 points) Infrared waves Microwaves Ultra

violet waves X-rays

1 answer:

Irina18 [472]3 years ago

5 0

In order from shortest to longest frequency, the electromagnetic waves can be ranked :

1. X-rays

2. Ultraviolet

3. Infrared

4. Microwave

X-rays has the shortest frequency which ranges between 30 petahertz to 30 exahertz. X-rays have the smallest wavelengths at 0.01-10 nanometers. Its energy ranges between 100eV to 100keV.

You might be interested in

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

Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th

Delvig [45]

Explanation:

As the given data is as follows.

$R_{1} = 680 \ohm$ ohm $\ohm$ , $R_{2} = 720 \ohm$ ohm,

$R_{3} = 1.2 k\ohm$ = 1200 $\ohm$ (as 1 k ohm = 1000 m)

(a) We will calculate the maximum resistance by combining the given resistances as follows.

Max. Resistance = $R_{1} + R_{2} + R_{3}$

= $(680 + 720 + 1200) \ohm$ ohm

= 2600 ohm

or, = 2.6 $k\ohm$ ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 $k\ohm$ ohm.

(b) Now, the minimum resistance is calculated as follows.

Min. Resistance = $\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

= $\frac{1}{680} + \frac{1}{720} + \frac{1}{1200}$

= $3.683 \times 10^{-3}$ ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is $3.683 \times 10^{-3}$ ohm.

3 0

2 years ago

Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C.

zaharov [31]

Answer:

c. is more than that of the fluid.

Explanation:

This problem is based on the conservation of energy and the concept of thermal equilibrium

$heat= m s \Delta T
$

m= mass

s= specific heat

\DeltaT=change in temperature

let s1= specific heat of solid and s2= specific heat of liquid

then

Heat lost by solid= $20(s_1)(70-30)=800s_1
$

Heat gained by fluid= $100(s_2)(30-20)=1000s_2
$

Now heat gained = heat lost

therefore,

1000 S_2=800 S_1

S_1=1.25 S_2

so the specific heat of solid is more than that of the fluid.

8 0

3 years ago

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago

Atmospheric conditions near a mountain range are such that a cloud at an altitude of 2.00 km contains 3.20 ✕ 10^7 kg of water va

kykrilka [37]

Answer:

time required is 6.72 years

Explanation:

Given data

mass m = 3.20 ✕ 10^7 kg

height h = 2.00 km = 2 × 10^3 m

power p = 2.96 kW =2.96 × 10^3 J/s

to find out

time period

solution

we know work is mass × gravity force × height

and power is work / time

so we say that power = mass gravity force × height / time

now put all value and find time period

power = mass × gravity force × height / time

2.96 × 10^3 = 3.20 ✕ 10^7 × 9.81× 2 × 10^3 / time

time = 62.784 × 10^10 / 2.96 × 10^3

time = 21.21081081 × 10^7 sec

time = 58.91891892 × 10^3 hours

time = 6.72 years

so time required is 6.72 years

3 0

3 years ago