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3 years ago

(ii) Electromagnetic waves transfer energy.

Physics

1 answer:

Andrews [41]3 years ago

8 0

Explanation:

6 examples

radio wave

infrared ray

visible light ray

ultraviolet ray

microwave

gamma ray

3 which are useful

infrared ray

radio waves

gamma ray

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Pluto has been reassigned and is now a dwarf planet. Why did scientists think this reassignment was necessary? If you were a sci

never [62]

Pluto was discovered by the astronomer Clyde Tombaugh in February 1930. It was given the status of the ninth planet of the solar system.
As telescopes, particularly in on satellites, improved, more objects were discovered which caused a problem that they were quite small and some astronomers didn't think they qualified as being planets.
The International Astronomical Union (IAU) had a vote which was very close. They defined three criteria which a planet must satisfy.
It must be large enough for gravity to overcome structures of materials and make it spherical. Most bodies are flattened spheroids due to rotation.
It must orbit the Sun.
It must have cleared its orbit of other bodies other than moons.
The IAU created a new definition of an object called a dwarf planet which only satisfies the first two criteria. Pluto fails the third criterion, so it was demoted to a dwarf planet.
Many people, including myself, still consider Pluto to be the ninth planet.
To be pedantic, Jupiter has a lot of asteroids in its orbit at its two Lagrange points. They are called trojan asteroids. So, this means that Jupiter fails the IAU's third criterion and should be a dwarf planet, which it is certainly not!

7 0

3 years ago

Read 2 more answers

How many complete wave cycles are shown?

skelet666 [1.2K]

Answer:

2.0

Explanation:

8 0

4 years ago

If everything that is created has to have an end, what could cause the end of the universe itself?

bearhunter [10]

Answer:

The Big Crunch hypothesis is a symmetric view of the ultimate fate of the universe. Just as the Big Bang started as a cosmological expansion, this theory assumes that the average density of the universe will be enough to stop its expansion and begin contracting.

Explanation:

hope it helps

5 0

3 years ago

And air rifle pellet of mass 2 g Is fired into a block of modelling clay mounted on a model railway track. The trunk and modelli

Anna007 [38]

Answer:

0.0816 kgm/s

Explanation:

From the question,

Momentum of the pellet just before it hits the modelling clay is = (mass of the pellet+ mass of the truck clay)×initial velocity of pellet.

P =(M+m)u...................... Equation 1

Where P = initial momentum of the pellet, m = mass of the pellet, u = initial speed of the pillet, M = mass of the truck

Given: m = 2 g = 2/1000 kg = 0.002 kg, u = 0.8 m/s, M = 0.1 kg

Substitute these values into equation 1

P = (0.002+0.1)0.8

P = (0.102)0.8

P = 0.0816 kgm/s

4 0

3 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

3 0

3 years ago