Remember your kinematic equations for constant acceleration. One of the equations is

, where

= final position,

= initial position,

= initial velocity, t = time, and a = acceleration.
Your initial position is where you initially were before you braked. That means

= 100m. You final position is where you ended up after t seconds passed, so

= 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was

= 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
Your acceleration is approximately
.
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
Answer:
In the case of a solution transition metal complex that has an absorption peak at 450 nm in the blue region of the visible spectrum, the (complementary) color of this solution is orange (option B).
Explanation:
The portion of UV-visible radiation that is absorbed implies that a portion of electromagnetic radiation is not absorbed by the sample and is therefore transmitted through it and can be captured by the human eye. That is, in the visible region of a complex, the visible color of a solution can be seen and that corresponds to the wavelengths of light it transmits, not absorbs. The absorbing color is complementary to the color it transmits.
So, in the attached image you can see the approximate wavelengths with the colors, where they locate the wavelength with the absorbed color, you will be able to observe the complementary color that is seen or reflected.
<u><em>
In the case of a solution transition metal complex that has an absorption peak at 450 nm in the blue region of the visible spectrum, the (complementary) color of this solution is orange (option B).</em></u>
Complete Question
Part of the question is shown on the first uploaded image
The rest of the question
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
Answer:
The net force exerted on the third charge is
Explanation:
From the question we are told that
The third charge is 
The position of the third charge is 
The first charge is 
The position of the first charge is 
The second charge is 
The position of the second charge is
The distance between the first and the third charge is


The force exerted on the third charge by the first is

Where k is the coulomb's constant with a value 
substituting values
The distance between the second and the third charge is


The force exerted on the third charge by the first is mathematically evaluated as
substituting values

The net force is
substituting values

Answer:
Option A; ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.
Explanation:
A red supergiant star is a larger and brighter type of red giant star. Red supergiants are often variable stars and are between 200 to 2,000 times bigger than the Sun. Example is Betelgeuse.
Betelgeuse is one of the largest known stars, it has a diameter of about 700 times the size of the Sun or 600 million miles, it emits almost 7,500 times as much energy as the Sun, it has a rather low surface temperature (6000F compared to the Sun's 10,000F); this means that it has a more cooler surface than the Sun's surface.
This low temperature also means that the star will appear orange-red in color, and the combination of size and temperature makes it a kind of star called a red super giant.
Although, all the statements above are correct, the only one that can be inferred from the red color of Betelgeuse is that ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.