Sorry Man If you ask this question no one wants to write an essay and waste their time. You are smart enough to write an essay.
Answer:
No.
Explanation:
The force that two particle experience is inversely proportional to the sqare of the distance, this is:
for a distance D
If we move them so that D is doubled:
= 
Then the force they experience is one fourth of the original.
Answer:
(4) weight
Explanation:
The centripetal force acting on the space shuttle in orbit is given by:
where
m is the mass of the shuttle
v is the tangential speed of the shuttle
r is the radius of its circular orbit
When the shuttle orbits the Earth, the centripetal force that keeps the shuttle in circular motion is given by the gravitational attraction between the shuttle and the Earth, which corresponds to the weight of the shuttle, and it is given by:
where
G is the gravitational constant
M is the Earth's mass
And this force, therefore, corresponds to the centripetal force.
In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because of friction.
AMA (Actual mechanical advantage) is found by dividing output force by effort force. The actual mechanical advantage will always be less than the ideal mechanical advantage. The ideal mechanical advantage assumes perfect efficiency which doesn't account for friction, while actual mechanical advantage does. Therefore; the IMA is always greater than the actual mechanical advantage because all machines must overcome friction.
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.
