Answer:
a) m₁ = m₂ F₁ₓ = F₂ₓ
b) m₁ << m₂ F₂ₓ =0
Explanation:
This interesting exercise is unclear your statement, so that in a center of mass system has an acceleration of zero it is necessary that the sum of the forces on each axis is zero, to see this we write Newton's second law
∑ F = m a
for acceleration to be zero implies that the net force is zero.
we must write the expression for the center of mass
[tex]x_{cm}[/tex] = 1 / M (m₁ x₁ + m₂ x₂)
now let's use the derivatives
[tex]a_{cm}[/tex] = d² x_{cm}/dt² = 1 / M (m₁ a₁ + m₂a₂)
where M is the total mass M = m₁ + m₂
so that the acceleration of the center of mass is zero
0 = 1 / M (m₁ a₁ + m₂a₂)
m₁ a₁ = - m₂ a₂
In the case that we have components on the x axis, the modulus of the two forces are equal and their direction is opposite, therefore
F₁ₓ = -F₂ₓ
b)r when the two masses are equal , in the case of a mass greater than the other m₁ << m₂
acm = d2 xcm / dt2 = 1 / M (m1 a1 + m2a2)
so that the acceleration of the center of mass is zero
0 = 1 / M (m1 a1 + m2a2)
m1 a1 = - m 2 a2
with the initial condition, we can despise m₁, therefore
0 = m₂a₂
if we use Newton's second law
F₂ = 0
I tell you that in this case with a very high mass difference the force on the largest mass must be almost zero
How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of it? How many times can something with a radius of 2,000 units fit something with a radius of 1 unit?
Answer:
# _units = 1000
Explanation:
This exercise we can use a direct proportion rule.
If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?
# _units = V₁₀ / V₁
The volume of a body of radius 1 is
V₁ = 4/3 π r₁³
V₁ = 4/3π
the volume of a body of radius r = 10
V₁₀ = 4/3 π r₂³
V10 = 4/3 π 10³
the number of times this content is
#_units = 4/3 π 1000 / (4/3 π 1)
# _units = 1000