Answer:
m = 35.98 Kg ≈ 36 Kg
Explanation:
I₀ = 125 kg·m²
R₁ = 1.50 m
ωi = 0.600 rad/s
R₂ = 0.905 m
ωf = 0.800 rad/s
m = ?
We can apply The law of conservation of angular momentum as follows:
Linitial = Lfinal
⇒ Ii*ωi = If*ωf <em>(I)</em>
where
Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m
If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m
Now, we using the equation <em>(I) </em>we have
(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800
⇒ m = 35.98 Kg ≈ 36 Kg
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;

where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

Therefore, the mutual force between the two point charges is 319.64 N
Decrease the amount of force applied
Answer:
a

b

Explanation:
From the question we are told that
The initial position of the particle is 
The initial velocity of the particle is 
The acceleration is 
The time duration is 
Generally from kinematic equation

=> 
=> 
Generally from kinematic equation

Here s is the distance covered by the particle, so

=> 
Generally the final position of the particle is

=> 
=> 