you can subtract the atomic number from the mass number to find the number of neutrons.
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
Answer:
- Fx = -9.15 N
- Fy = 1.72 N
- F∠γ ≈ 9.31∠-10.6°
Explanation:
You apparently want the sum of forces ...
F = 8.80∠-56° +7.00∠52.8°
Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...
f∠α = (-f·cos(α), -f·sin(α))
This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.
= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))
≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)
≈ (-9.15309, 1.71982)
The resultant component forces are ...
Then the magnitude and direction of the resultant are
F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)
F∠γ ≈ 9.31∠-10.6°
Answer:
the force between the building and the ball is non-conservative (friction-type force)
Explanation
Explanation:For this exercise the student must create an impulse to move the ball towards the building, in this part he performs positive work since the applied force and the displacement are in the same direction.
When the ball moves it has a kinetic energy and if its height increases or decreases its potential energy also changes, but the sum of being must be equal to the initial work.
When the ball arrives and collides with the building, non-conservative forces, of various kinds; rubbing, breaking, etc. It transforms this energy into a part of heat and another in mechanical energy that the building must absorb, let us destroy its wall
Consequently, the force between the building and the ball is non-conservative (friction-type force
