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4 years ago

What is light? Makes thing visible?

1 answer:

6 0

Yes, light enters your eye and makes things visible. It's only a small portion of the elctromagnetic spectrum. It's called "visible light." What other things does the question want to know? I could go in a describe what light really is but I think itd be unnecessary.

You might be interested in

What do counties get their power from

SOVA2 [1]

Money, or dictatiorship, trading, something that involes with money

8 0

4 years ago

What term refers to substance in the environment that can cause abnormal fetal development

sveta [45]

Answer:

Teratogens

Explanation:

Teratogens are substances or other factors that can cause congenital abnormalities, which are also called birth defects. Usually these abnormalities arise in the third to eighth weeks of pregnancy, when the major organ systems are forming.

6 0

3 years ago

At an amusement park, a swimmer uses a water slide to enter the main pool. a. If the swimmer starts at rest, slides with negligi

adoni [48]

Answer:

a)6.7m/S

b)6.8m/s

Explanation:

Hello ! To solve the point b you must follow the steps below

1.Draw the slide taking into account its length and height and find the angle from which the swimmer is launched (see attached image)

2. Find the horizontal velocity (X) and vertical (Y) components (see attached image)

3) for the third step we must remember that as in the slide there is no horizontal acceleration the speed in X will remain constant at the end of the swimmer's path (Vx = 0.59m / s)

the fourth step is to remember that vertically there is constant acceleration called gravity (g = 9.81m / s ^ 2), so to find the speed at the end of the route we use the following equation

$Vfy= \sqrt{Vy^2+2gy}$

where

Vfy= final verticaly speed

Vy=initial verticaly speed=0.59m/S

g=gravity=9.81m/S^2

y=height of slide=2.31m

solving

$Vfy= \sqrt{Vy^2+2gy}\\Vfy= \sqrt{(0.59)^2+2(9.81)(2.31)}=6.77m/s$

The last step is to add the velocity components vectorally at the end of the route with the following equation

$V=\sqrt{Vfy^2+Vx^2} =\sqrt{6.77^2+0.59^2} =6.8m/s$

point A

taking into account the previous steps we can infer that as the swimmer starts from rest, the velocity (Vx=Vy=O) is zero, so we should only use the formula for constant acceleration movement.

$Vfy= \sqrt{Vy^2+2gy}$

vy=0

$Vfy=\sqrt{2gy}$

Vfy= $\sqrt{2(9.81)(2.31)}$ =6.7m/s

8 0

4 years ago

What would be the weight of a 59.1-kg astronaut on a planet with the same density as Earth and having twice Earth's radius?

Olegator [25]

You could always directly use F = G M m / r², but that takes forever.

It is best to know that, on Earth's surface, the astronaut would weigh 
F = m g = 579.2 N 
where m = 59.1 kg and g = 9.8 m/s². 

You then notice from F = G M m / r² that doubling the mass of the earth would double his weight and doubling the radius would give 1/4 his original weight. So... 
F = 1/4 * 2 * 579.2 N = 290 N

5 0

3 years ago

Consider the following three statements: For any electromagnetic radiation, the product of the wavelength and the frequency is a

boyakko [2]

Answer:

The correct answer is B. Statements (i) and (ii) are true.

Explanation:

Fisrt statement:

Electromagnetic radiation such as wave, wavelength λ and oscillation frequency ν are related by a constant, the speed of light. The equation is given by:

$c= \lambda \nu$

So the first statement is true.

Second statement:

$c=\lambda \nu\\c= 3.0 A \times 1.0\times 10^{18} Hz\\c= 3.0\times 10^{-10} m \times 1.0\times 10^{18} s^{-1}\\c=3\times 10^8 \frac{m}{s}$

The value of c in the vacuum is 3×10⁸ m/s. Hence, the second statement is true.

Third statement:

The speed of any electromagnetic radiation is constant regardless the type of radiation.

Hence, the third statement is false.

8 0

3 years ago