Mass of yellow train, my = 100 kg
Initial Velocity of yellow train, = 8 m/s
mass of orange train = 200 kg
Initial Velocity of orange train = -1 m/s (since it moves opposite direction to the yellow train, we will put negative to show the opposite direction)
To calculate the initial momentum of both trains, we will use the principle of conservation of momentum which
The sum of initial momentum = the sum of final momentum
Since the question only wants the sum of initial momentum,
(100)(8) + (200)(-1) = 600 m/s
In this problem,
Applied force(F) = 10 N
The object’s mass (m) is 5 kg.
Having said that,
An object’s force is equal to the product of its mass and the acceleration it experiences as a result of the applied force.
i.e., Mass + Acceleration = Force (a)
F= m×a
Therefore,
A= F÷m
A= (10÷5) m/sec²
A= 2 m/sec²
Consequently, the object’s acceleration,
A=2 m/sec²
Concept of force and acceleration:
This states that the rate of velocity change of an object is directly proportional to the applied force and moves in the direction of the applied force.
It can be expressed mathematically as force (N) = mass (kg) x acceleration (m/s2). Therefore, an object with constant mass will accelerate in direct proportion to the applied force.
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normal force=mass*gravitational force
normal force=40*0
normal force=40
Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis
![F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}](https://tex.z-dn.net/?f=F_N%3Dk_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D)
Explanation:
Let the force on +Q charge y-axis due to +2Q charge be 
and force on +Q charge y axis due to -Q charge on x-axis be 
.
Distance between the +2Q charge and +Q charge = d units
Distance between the -Q charge and +Q charge = 
units
= Coulomb constant
Net force on +Q charge on y-axis is:
![|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|](https://tex.z-dn.net/?f=%7CF_N%7C%3D%7Ck_e%5Cfrac%7BQ%5E2%7D%7Bd%5E2%7D%5Ctimes%20%5Csqrt%7B%5B4%2B%5Cfrac%7B1%7D%7B4%7D-%5Csqrt%7B2%7D%5D%7D%7C)
The net froce on the +Q charge on y-axis is
