Answer:
485520 m
Explanation:
= initial velocity of the projectile = 1360 m/s
= final velocity of the projectile =
=
= 544 m/s
a = acceleraton due to gravity on moon = - 1.6 m/s²
h = Altitude of the projectile
Using the kinematics equation
Inserting the values
h = 485520 m
Answer:
(a) The average speed is 0.85 milles/minute
(b) The average velocity is zero
Explanation:
In order to answer part (a) and (b) you have to apply the formulas for average speed and average velocity which are:
<em>-Average speed formula:</em>
where d is the total distance traveled and t is the tota time
Replacing the given values:
milles/minute
Notice that you have to replace the total distance, which is 14 milles for the go plus 14 milles for the return. The same for the total time.
<em>-Average velocity formula:</em>
V = Δx/Δt
Where V is the velocity vector, Δx is the displacement and Δt is the change in time
V=
Where X2 is the final position and X1 is the initial position
In this case X1= 0 i and X2=0 i (i is the unit vector in the x direction). So, the displacement is zero.
Therefore, the average velocity is:
V= 0 i [milles/minute]
Answer:
= 4
Explanation:
To solve this exercise we will use Bohr's atomic model
= - 13.606 / n² [eV]
The transition from level n = 2 to level n = 1 is valid
= - 13.606 [¼ -1/1]
= 10.2045 eV
Bohr's model for atoms with only one electron is
= -13.606 Z² / n²
Where Z is the atomic number of the atom.
In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches
ΔE = -13.606 [4 / ² - 4/4]
4 / ² = -ΔE / 13.606 + 1
4 / ² = -10.2045 / 13.606 +1 = -0.75 +1
4 / ² = 0.25
= √ 4 / 0.25
= 4
1) Do > 2f
Do means object distance
2f means 2 x focal length
2 x focal length = radius of curvature
When an object is placed beyond the radius of curvature, a real image is formed between the radius of curvature and focus. The image size is reduced. The sketch is shown below
