In a fluorescent tube of diameter 3 cm, 3 1018 electrons and 0.75 1018 positive ions (with a charge of e) flow through a cross-s
1 answer:
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Explanation:
What is the weight of a 2.00-kilogram object on the surface of Earth?
2.00 N
4.91 N
9.81 N
19.6 N
Given parameters:
Mass of the object = 2kg
Unknown:
Weight of the object = ?
Solution:
The weight of an object is the force of gravity acting on the object;
Weight = mass x acceleration due to gravity
Acceleration due to gravity = 9.8m/s²
Now insert the parameters and solve;
Weight = 2 x 9.8 = 19.6N
A person weighing 785 Newtons on the surface of the Earth would weigh 47 Newtons on the surface of Pluto. What is the magnitude of the gravitational acceleration on the surface of Pluto?
1.7 m/s²
0.59 m/s²
0.29 m/s²
9.8 m/s²
Given parameters:
Weight on Earth = 785N
Weight on Pluto = 47N
Unknown:
Acceleration due to gravity on Pluto = ?
Solution
The mass of the body both on Earth and Pluto is the same.
Weight = mass x acceleration due to gravity
Now find the mass on Earth;
Acceleration due to gravity on Earth = 9.8m/s²
785 = mass x 9.8
mass = = 80.1kg
So;
Acceleration due to gravity on Pluto =
Acceleration due to gravity = = 0.59m/s²
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
= 1 / M (m₁ x₁ + m₂ x₂)
now let's use the derivatives
= 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