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nordsb [41]
3 years ago
6

Could someone please explain what makes up the electromagnetic spectrum? Thank you.

Physics
2 answers:
Sonja [21]3 years ago
4 0
Radio waves, microwaves, infrared waves, visible light, ultraviolet, x-rays, and gamma rays.

-Waves can travel through a vaccum
-These waves radiate in all directions
-travel through speed of light
-they are transverse waves
-they are made by interaction between an rlectric and a magnetic field.

These are the main points. I hope this will help you! ❤️.
faust18 [17]3 years ago
3 0
These types include gamma rays, x-rays, ultraviolet, infrared, microwaves and radio waves. Together with visible light, all these types of radiation make up what we call the electromagnetic spectrum - the complete spectrum of radiation.
You might be interested in
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
How do I calculate velocity
Jobisdone [24]

Answer: See photo

Explanation: There are a couple of ways to use velocity in an equation in the photo.

5 0
3 years ago
The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

8 0
3 years ago
Read 2 more answers
Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis.
amm1812

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

8 0
3 years ago
A 100kg crate slides along a floor with a starting velocity of 21ms If the force due to friction is 8N how long will it take for
GaryK [48]

A 100kg crate slides along a floor with a starting velocity of 21 m/s. If the force due to friction is 8N, then, it will take 262.5 s for the box to come to rest.

We'll begin by calculating the declaration of the box. This can be obtained as follow:

Force (F) = –8 N (opposition)

Mass (m) = 100 Kg

<h3>Deceleration (a) =? </h3>

<h3>F = ma</h3>

–8 = 100 × a

Divide both side by 1000

a = \frac{-8}{100}

<h3>a = –0.08 ms¯²</h3>

Therefore, the deceleration of the box is –0.08 ms¯²

Finally, we shall determine the time taken for the box to come to rest. This can be obtained as follow:

Deceleration (a) = –0.08 ms¯²

Initial velocity (u) = 21 ms¯¹

Final velocity (v) = 0 ms¯¹

<h3>Time (t) =.? </h3>

<h3>v = u + at</h3>

0 = 21 + (–0.08×t)

0 = 21 – 0.08t

Collect like terms

0 – 21 = –0.08t

–21 = –0.08t

Divide both side by –0.08

t = \frac{-21}{-0.08}\\\\

<h3>t = 262.5 s</h3>

Therefore, it will take 262.5 s for the box to come to rest.

Learn more: brainly.com/question/14446351

7 0
2 years ago
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