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3 years ago

List the parts of the electromagnetic spectrum from longest wavelength to shortest wavelength

1 answer:

6 0

The rays of the electromagnetic spectrum from shortest to longest wavelength are: radio waves, microwaves, infrared rays, optical rays, ultraviolet rays, X-rays, and gamma-rays.

EDIT: He has these backwards, the shortest wavelength is created by Gamma-Rays and the longest is Radiowaves.

 Remember- high energy = short wavelength.

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The voltage supplied to a circuit is 17 V and the current running through is 10 A. What is the power generated?

Lelu [443]

Answer:

170 W

Explanation:

Applying

P = VI.................... Equation 1

Where P = Power generated in watt, V = Voltage supplied to the circuit, I = Current running through the circuit.

From the question,

Given: V = 17 V, I = 10 A

Substitute these values into equation 1

P = (17×10)

P = 170 Watt.

Hence the power generated is 170 W.

The right option is A. 170 W

3 0

2 years ago

Un vas plin cu lichid cântăreşte 175kg. Ceea ce reprezintă de 5 ori masa vasului gol. Ştiind că volumul interior al vasului este

Mariana [72]

a) Density of the liquid: $823.5kg/m^3$

b) Weight of the liquid: 1372 N

Explanation:

Translation of the text:

"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate:

a) the density of the liquid;

b) the weight of the liquid."

We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as

$M=m_L + m_V$ (1)

where

$m_L$ is the mass of the liquid

$m_V$ is the mass of the vessel

We also know that the total mass M is 5 times the mass of the empty vessel, so we have:

$M=5m_V\\m_V=\frac{M}{5}=\frac{175}{5}=35 kg$

which is the mass of the empty vessel.

Therefore, we can find the mass of the liquid only using (1):

$m_L=M-m_V=175-35=140 kg$

The density of the liquid is given by

$d=\frac{m}{V}$

where

m = 140 kg (mass of the liquid)

V = 0.170 kL = 170 L = $0.170 m^3$ (volume of the liquid, which is equal to the volume of the vessel)

So we get

$d=\frac{140}{0.170}=823.5kg/m^3$

The weight of a body is given by

$F=mg$

where

m is its mass

g is the acceleration due to gravity

For the liquid in this problem, we have

m = 140 kg (mass)

$g=9.8 m/s^2$ (acceleration due to gravity)

Therefore, its weight is

$F=(140)(9.8)=1372 N$

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

6 0

3 years ago

What type of change accurs when a substance stays the same

Thepotemich [5.8K]

This would be a physical change because it can change back to its original form. This is like ripping paper. You can piece it back together and it still is paper.

The opposite of this is chemical change. Chemical change means the product has been changed completely like burning paper. The paper has now been turned to ash and it's impossible to change this back to its original form.

3 0

3 years ago

In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the

Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

8 0

3 years ago

A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0

3 years ago