Answer: 
Explanation:
The equation to calculate the center of mass 
of a particle system is:
In this case we can arrange for one dimension, assuming the geometric center of the Earth and the ladder are on a line, and assuming original center of mass located at the Earth's geometric center:
Where:
is the mass of the Earth
is the mass of 1 billion people
is the radius of the Earth
is the distance between the center of the Earth and the position of the people (2 m above the Earth's surface)
This is the displacement of Earth's center of mass from the original center.
Answer:
Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is
The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
![\displaystyle F_a=F_{r1}+F_{r2}.....[1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_a%3DF_%7Br1%7D%2BF_%7Br2%7D.....%5B1%5D)
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is
The friction forces are computed by
Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]
Simplifying
Plugging in the values
![\displaystyle F_{a}=0.25(9.8)[400+2(100)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_%7Ba%7D%3D0.25%289.8%29%5B400%2B2%28100%29%5D)
Answer:
I believe the answer is Plasma
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
