Answer:
a. 
b. 
Explanation:
The inertia can be find using
a.





now to find the torsion constant can use knowing the period of the balance
b.
T=0.5 s

Solve to K'


Answer:
When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u> m.
When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is<u> 0.43</u> m.
When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is <u>0.87</u> m.
When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is <u>1.25</u> m.
When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is <u>1.86</u> m.
Sorry for not answering early on! If anyone in the future needs help, I got these answers from 2020 egenuity, though I can't post the picture for proof. Stay Safe!
Answer:

Explanation:
You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:


where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.
by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.
By doing the same for the electron you obtain:

we can equals these expressions for both proton and electron, because the forces qE are the same:

Answer:
20 cm
Explanation:
Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².
So r = kq₁q₂/U
x - 2 = kq₁q₂/U
x = 0.02 + kq₁q₂/U m
x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J
x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J
x = 0.02 + 0.18 = 0.2 m = 20 cm
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>