Answer:
1. Least massive stars are the coolest and least luminous, lower right of main sequence, on HR diagram.
2. Most massive are the hottest and most luminous, upper left of main sequence on Hr Diagram.
3. The radius of stars are related to their sprectral type. having the O being the hottest upper left and M being the coolest bottom right.
1. What is the major function of the pons?
-Other than serving as a conduction pathway, the pons also contains a nucleus that is essential for breathing.
2. Why is the medulla the most vital part of the brain?
-Medulla is very important because it function for the involuntary movement and processes that happen in the body without giving a thought about it. It performs crucial tasks like regulating the blood pressure.
Answer: The hottest star is Archenar( blue) and the coolest star is Betelgeuse
Explanation:
Objects emit radiation that depends exclusively on their temperature. At an ambient temperature, the radiation emitted by an object is in the infrared spectrum (we could only see it with a special camera). If we heat it we will see that it first turns red (whose state we call “red hot”) because it is the lowest and least energetic wavelength of all.
If we continue to heat it, the wavelength that it emits to one with more energy will continue to increase and we will see that it turns yellow and then white. This is a signal that is emitting at all frequencies (but mainly in blue).
If we continue to warm a body that is "white hot", it would emit in the ultraviolet spectrum, with what would become ... black! then we would not see it emits light in the visible spectrum (well, we would see a very faint bluish light corresponding to the tail of the distribution of the spectrum it emits, but the peak of that spectrum would be in the ultraviolet).
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake