Sedimentary rock I believe
The question is poor. Light doesn't refract on its way THROUGH anything. It refracts at the boundary BETWEEN two different media. The effect is greatest where the ratio of the speeds of light in the two media is greatest. On your list, that would be at the boundary between air or space and glass.
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.
We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.
The values of the velocities are,
We know that the acceleration is equivalent to the change of the speed in a certain time therefore
Now applying the Newton's second law we have,
Therefore the approximate magnitude is 8516.36N
Answer:
a. 11 m/s at 76° with respect to the original direction of the lighter car.
Explanation:
In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:
Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.
The magnitude of the final velocity of the wreck can be found as:
![v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bfx%7D%5E%7B2%7D%2B%20v_%7Bfy%7D%5E%7B2%7D%5C%5Cv_%7Bf%7D%3D%5Csqrt%5B%5D%7B2.6%5E%7B2%7D%20%2B%2010.4%5E%7B2%7D%7D%20%5C%5Cv_%7Bf%7D%3D%2010.72)
The final velocity has an intensity of roughly 11 m/s
As for the angle, it can be determined in respect to the lighter car (x axis) as follows:
Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.
