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3 years ago

In what regions of the electromagnetic spectrum is the atmosphere transparent enough to allow observations from the ground?

Physics

1 answer:

MissTica3 years ago

3 0

Answer:

Visible Light and Radio waves

Explanation:

The earth's atmosphere is transparent to a few windows in the electromagnetic spectrum. it is completely transparent to allow observation from the ground in visible light rang 380 to 740 nano meters. Also in the range of radio wave as communication are done from space to ground in the form of radio waves.

it is Partially transparent to Microwave and infrared range.

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

2 years ago

1 Write 7 next to the statements that are true.

Grace [21]

Answer: The pressure in a liquid dec reaches with depth. F

The pressure in a liquid increases with depth.

The upthrust on an object is larger when it is deeper in a pool. 7

The bottom of a dam is thinner than the top of a dam. F

The bottom of a dam is thicker than the top of a dam.

The pressure is bigger at the bottom of a lake because of the weight of water above it. 7

I think these are the answers.

7 0

3 years ago

Find the volume of the irregular object a marble

Lostsunrise [7]

By using Displacement method we can find volume of irregular object like marble.

<h3>What is displacement method?</h3>

In displacement method,

First , we measuring the volume of water displaced by an object which tell us the volume of the object.

Secondly We can use the physical balance to determine its mass.

Lastly , calculate the density by dividing the mass by the volume.

Procedure to find the Volume of irregular object i.e. marble

step 1 - Fill the graduated cylinder about half full and measure the initial volume of water.

step 2 - Drop the marble in the graduated cylinder.

step 3 -Now measure the final level of water.

Step 4 - Subtract both the values.

Here , comes the Volume of Irregular Object.

For more volume related question visit here:

brainly.com/question/1578538

#SPJ1

7 0

1 year ago

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car?

aksik [14]

ANSWER

C. $F=4.5 \times10^3$ . newtons

EXPLANATION

According to Newton's second law,

$F_{net}=ma$ , where

$m=1.00\times 10^3kg$ is the mass measured in kilograms.

and

$a=4.5ms^{2}$ is the acceleration in metres per second square.

We substitute these values to obtain,

$F=1.00\times10^3 \times 4.5$ .

We rearrange to get,

$F=1.00\times4.5 \times10^3$ .

We multiply out the first two numbers and leave our answer in standard form to get,

$F=4.5 \times10^3 N$ .

The correct answer is C

3 0

3 years ago

A student used a yardstick to model the displacement of rock at a fault during an earthquake. The student bent the yardstick wit

jonny [76]

I think the correct answer would be B. The process of elastic rebound is being shown by the student. It is a theory that is used to explain earthquakes. It focuses on how energy is being spread in times of earthquakes. As the rocks on the fault experiences shift and force, these rocks would be accumulating energy causing it to deform reaching the internal strength and eventually exceeding it. At that moment, a rapid motion would happen along the fault, which releases the energy, then the rocks would go back to its original shape or the undeformed state. This theory is the first theory that sufficiently was able to explain earthquakes.

7 0

3 years ago

Read 2 more answers