All categories

3 years ago

In which region of the electromagnetic spectrum is this radiation found?

1 answer:

6 0

<span>The electromagnetic spectrum consists of all the different wavelengths of electromagnetic radiation, including light, radio waves, and X-rays.
</span>The only region in the entire electromagnetic spectrum that our eyes are sensitive to is the visible region.

You might be interested in

Which statement describes something that should always be done at the end of a calculation?

Reika [66]

Answer:

C) Check that the numerical answer is reasonable and the units are what you expected.

Explanation:

5 0

3 years ago

Read 2 more answers

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. it suddenly collides directly with a stationary seal of m

Sphinxa [80]

M = mass of the whale = 1000 kg

m = mass of the seal = 200 kg

V = initial velocity of whale before collision with the seal = 6.0 m/s

v = initial velocity of the seal before collision with the whale = 0 m/s

V' = final velocity of two sea creatures after collision = ?

Using conservation of momentum

M V + m v = (M + m) V'

inserting the above values in the equation

(1000 kg) (6.0 m/s) + (200 kg) (0 m/s ) = (1000 kg + 200 kg) V'

6000 kgm/s + 0 kgm/s = (1200 kg) V'

V' = (6000 kgm/s ) /(1200 kg)

V' = 5 m/s

3 0

3 years ago

Read 2 more answers

Two uniform spheres,each of mass 0.260kg are fixed at points A and B

Ivan

Answer:

a) 2.76x10^-9 m/s^2

b) downward

Explanation:

a) In order to do this, we need to use the newton's law of gravitation which is:

F = G.m/r²

Where:

G: Constant of gravitation (6.67x10^-11 N/m)

m: mass of the object

r: distance or radius of the object.

In this case, we have two spheres, so this law becomes:

F = G*m1*m2 / r²

Now, you are not providing a picture with this, however, this problem is very similar to one I solved before, so, I'm gonna use the same values to solve this. If, you have different values for the distance, just replace them in this procedure and should get the correct and more accurate answer.

In this case, the distance that separates sphere in P from sphere A and B is 10 cm. As both spheres have the same mass and they are at the same distance from P, we can assume that F1 = F2 so:

F1 = (6.67x10^-11) * (0.01) * (0.26) / (0.1)² = 1.73x10^-11 N

sphere at P, will roll downward to the middle of sphere A and B. The distance between these two points it's 8 cm (See picture attached)

Now, according to my data, they are exerting a force in the y axis, so:

F1y = Fy * senФ

senФ using trigonometry is:

senФ = opposite lenght / hipotenuse

senФ = 8/10 = 0.8

Replacing:

F1y = 1.73x10^-11 * 0.8 = 1.38x10^-11 N

This force it's exactly the same as F2y, so the total force is:

F = 1.38x10^-11 * 2 = 2.76x10^-11 N

Finally, using the second law of Newton:

F = m*a

Solving for a:

a = F/m

replacing:

a = 2.76x10^-11 / 0.01

a = 2.76x10^-9 m/s²

According to the picture and the value of this magnitude, it's going downward.

3 0

3 years ago

Place the types of radiation in order of INCREASING mass.

lubasha [3.4K]

Answer & explanation:

Gamma Radiation (γ): Also called gamma rays. Gamma radiation is made of electromagnetic waves, has zero charge and mass, continuously emits heat and has the ability to ionize air and make it conductive of electric current.

Beta radiation (β): Beta rays or beta particles, are electrons, negative particles with charge - 1 and mass number 0.
Alpha radiation (α): Also called alpha particles or alpha rays, are particles charged by two protons and two neutrons. They have positive charge +2 and mass number 4.

5 0

3 years ago

What should be the minimum thickness of this film?

NikAS [45]

Answer:

$d=93.75\,nm$ is the minimum thickness of the coating of given material.

Explanation:

This method of application of non reflective coatings uses the phenomenon of destructive interference in which the the waves reflected from the top surface of the coating are out of phase to the waves reflected from the bottom layer of the coating.

Given that:

refractive index of coating, $n_c=1.28$