Newton’s first law is commonly stated as:
An object at rest stays at rest and an object in motion stays in motion.
However, this is missing an important element related to forces. We could expand it by stating:
An object at rest stays at rest and an object in motion stays in motion at a constant speed and direction unless acted upon by an unbalanced force.
By the time Newton came along, the prevailing theory of motion—formulated by Aristotle—was nearly two thousand years old. It stated that if an object is moving, some sort of force is required to keep it moving. Unless that moving thing is being pushed or pulled, it will simply slow down or stop. Right?
This, of course, is not true. In the absence of any forces, no force is required to keep an object moving. An object (such as a ball) tossed in the earth’s atmosphere slows down because of air resistance (a force). An object’s velocity will only remain constant in the absence of any forces or if the forces that act on it cancel each other out, i.e. the net force adds up to zero. This is often referred to as equilibrium. The falling ball will reach a terminal velocity (that stays constant) once the force of air resistance equals the force of gravity.
Hope this help
Explanation:
Given that,
Charge 1, 
Charge 2, 
Distance between charges, r = 0.0209 m
1. The electric force is given by :
F = -492.95 N
2. Distance between two identical charges, 
Electric force is given by :
Hence, this is the required solution.
<span>Answer:
Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by
h = 0.5*g*T^2
However, the speed of the package-rocket system must be sufficient to cross the river in that time
v2 = L/T
Conservation of momentum says that
m1*v1 = (m1 + m2)*v2
where m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system.
Expressing v2 in terms of v1
v2 = m1*v1/(m1 + m2)
and then expressing the time in terms of v1
T = (m1 + m2)*L/(m1*v1)
substituting T in the first expression
h = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2
solving for v1, the speed before impact is given by
v1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1</span>
Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
