Answer:
the graph would lower down if we are talking about speed because the velocity stops going and theres no more kinetic energy to cause any speed
Explanation:
Answer:
(ω₁ / ω₂) = 1.9079
Explanation:
Given
R₁ = 3.59 cm
R₂ = 7.22 cm
m₁ = m₂ = m
K₁ = K₂
We know that
K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²
if
v₁ = ω₁*R₁
and
I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²
∴ K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² <em>(I)</em>
then
K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²
if
v₂ = ω₂*R₂
and
I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²
∴ K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂² <em>(II)</em>
<em>∵ </em>K₁ = K₂
⇒ 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²
⇒ ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²
⇒ (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²
⇒ (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)
⇒ (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²
⇒ (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)
⇒ (ω₁ / ω₂) = 1.9079
The center of mass isn't affected by the explosion.
To find the answer, we need to know about the trajectory of motion at zero external force.
<h3>How is the trajectory of an object changed when the net external force on it is zero?</h3>
- When there's no net external force acting on an object, its momentum doesn't change with time.
- As its momentum doesn't change, so it continues with the original trajectory.
<h3>Why doesn't the trajectory of firework change when it's exploded?</h3>
- When a firework is exploded, its internal forces are changed, but there's no external force.
- So, although the fragments follow different trajectories, but the trajectory of center of mass remains unchanged.
Thus, we can conclude that the center of mass isn't affected by the explosion.
Learn more about the trajectory of exploded firework here:
brainly.com/question/17151547
#SPJ4
Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F = 
q = 
we calculate
q = 
q = 
Ra 7 10-12
q = 2.65 10⁻⁶ C
