The orientation in space of an atomic orbital is associated with
1 answer:
<h2>Answer: The magnetic quantum number </h2><h2></h2>
The magnetic quantum number is one of the four quantum numbers that characterize the quantum state of an electron bound from an atom.
This number determines the spatial orientation of the orbital and its name is due to the fact that this spatial orientation is normally defined in relation to an external magnetic field.
Therefore:
<h2>The orientation in space of an atomic orbital is associated with <u>the magnetic quantum number</u></h2>
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Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
The compression of 10 cm by a 100 N force on the plane that has a
coefficient of friction of 0.39 give the following values.
- The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s
- The height at which the block stops rising is approximately 1.1415 m
- The length of the incline is approximately 1.536 m
Mass of the block, m = 3 kg
Coefficient of kinetic friction, = 0.39
Therefore, we have;
Friction force = ·m·g·cos(θ)
Which gives;
Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68
Work done by the motion of the block, <em>W</em> ≈ 7.68 × d
The work done = The kinetic energy of the block, which gives;
The initial kinetic energy in the spring is found as follows;
K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J
The initial velocity of the block is therefore;
5 = 0.5·m·v²
v₁ = √(2 × 5 ÷ 3) ≈ 1.83
Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J
Chane in kinetic energy, ΔK.E. = Work done
ΔK.E. = 0.5 × 3 × (v₁² - v₂²)
Which gives;
ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376
Which gives;
- The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>
The height at which the block will stop moving, <em>h</em>, is given as follows;
Which gives;
The distance up the inclined, the block rises, at maximum height is therefore;
≈ 1.536 m
Therefore;
h = 1.536 × sin(48°) ≈ 1.1415
- The height at which the block stops rising, h ≈ <u>1.1415 m</u>
From the above solution for the height, the length of the incline is he
distance along the incline at maximum height which is therefore;
- Length of the incline,
= 1.536 m
Learn more about conservation of energy here: