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2 years ago

The wave that carrys the most energy are the waves that?

2 answers:

5 0

"The different types of radiation are defined by the the amount of energy found in the photons. Radio waves have photons with low energies, microwave photons have a little more energy than radio waves, infrared photons have still more, then visible, ultraviolet, X-rays, and, the most energetic of all, gamma-rays."

Charra [1.4K]2 years ago

5 0

Don’t need a medium which is electromagnetic waves they can travel through a vacuum. Hope I helped

You might be interested in

Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object

ANEK [815]

Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem. Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them. So let's call them F1 and F2, with F1 arger than F2. Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.

Can you solve this system of equations seeing them like this, or do you need more help?

6 0

2 years ago

What is 12 and 1 half % expressed as a fraction

kiruha [24]

12 1/2 i would assume

6 0

3 years ago

Read 2 more answers

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

Your Lead Teaching Assistant is initially seated on the top of a hemispherical ice mound of radius R = 30 m. Approximate the ice

saw5 [17]

Answer:

Height = 5.65 km

Explanation:

$3\pi r^2$ is the circumference or we can say measures the boundary of hemisphere of friction-less ice that he is sitting on.

So, the height will be = 2 x 3.14 x (30)^2 = 5654.7 m = 5.65 km

7 0

3 years ago

Read 2 more answers

Despite a very strong wind, a tennis player

Gnoma [55]

Answer:

Option 5. 1 and 3

Solution:

The only forces acting on the tennis ball after it has left contact with the racquet and the instant before it touches the ground are the force of gravity in the downward direction and the force by the air exerted on the ball.

The ball after it left follows the path of trajectory and as it moves forward in the horizontal direction the force of the air acts on it.

In the whole projectile motion of the ball, the acceleration due to gravity acts on the ball thus the force of gravity acts on the ball in the downward direction before it hits the ground.

6 0

3 years ago