Explanation:
Red, green, and blue are therefore called additive primaries of light. ... When you block two lights, you see a shadow of the third color—for example, block the red and green lights and you get a blue shadow. If you block only one of the lights, you get a shadow whose color is a mixture of the other two.
First, your definition of a shadow is incorrect. A shadow is an area that receives less light than its surroundings because a specific source of light is blocked by whatever is "casting" the shadow. Your example of being outside reveals this. The sky and everything around you in the environment (unless you are surrounded by pitch black buildings) is sending more than enough light into your shadow, to reveal the pen to your eyes. The sky itself diffuses the sunlight everywhere, and the clouds reflect plenty of light when they are not directly in front of the Sun.
If you are indoors and have two light bulbs, you can throw two shadows at the same time, possibly of different darknesses, depending on the brightness of the light bulbs.
It can take a lot of work to get a room pitch black. One little hole or crack in some heavy window curtains can be enough to illuminate the room. There are very few perfectly dark shadows.
Answer:
P=6.25N and Q=16.25N
Explanation:
In order to solve this problem we must first draw a free body diagram for both situation, (see attached picture).
Now, we need to analyze the two free body diagrams. So let's analyze the first diagram. Since the body is accelerated, then the sum of forces is equal to mass times acceleration, so we get:
We can assume there will be only the two mentioned forces P and Q, so
the sum of forces will be:
P+Q=ma
P+Q=22.5N
We can do the same analysis for the second free body diagram:
Q-P=10.5N
so now we have a system of equations we can solve by elimination:
Q+P=22.5N
Q-P=10.5N
Now, we can add the two equations together so the P force is eliminated, so we get:
2Q=32.5N
now we can solve for Q:
so
Q=16.25N
Now we can use any of the equations to find P.
Q+P=22.5N
P=22.5N-Q
when substituting for Q we get:
P=22.5N-16.25N
so
P=6.25N
<span>Electric field is proportional to q/d^2, where q is the magnitude of the charge and d is the distance. Since all the given units are identical, we can just compare their relative magnitudes without calculating for the exact values.
A) 3/(0.4)^2 = 18.75
B) 1.5/(0.2)^2 = 37.5
C) 6/(0.4)^2 = 37.5
D) 3/(0.2)^2 = 75
Therefore, choice D has the largest electric field of all.
</span>
<span>After an exoplanet has been identified using a given detection method, scientists attempt to identify the basic properties of the planet which can tell us what it might be made of, how hot it might be, whether or not it contains an atmosphere, how that atmosphere might behave, and finally, whether the planet may be suitable for life. It is often useful to first determine basic properties of the parent star (such as mass and distance from the Earth). This is then followed by the use of planetary detection methods to calculate planetary mass, radius, orbital radius, orbital period, and density. The density calculation will provide clues as to what the planet is made of and whether or not it contains a significant atmosphere.
Mass and Distance of Parent Star
The mass and distance of an exoplanet's parent star must often be calculated first, before certain measurements of the exoplanet can be made. For example, determining the star's distance is an important step in determining a star's mass (see below). Knowing the mass of a star then allows the mass of the planet to be measured, for example when using the Radial Velocity Method.</span>
Answer: m∠P ≈ 46,42°
because using the law of sines in ΔPQR
=> sin 75°/ 4 = sin P/3
so ur friend is wrong due to confusion between edges
+) we have: sin 75°/4 = sin P/3
=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16
=> m∠P ≈ 46,42°
Explanation:
