The electrostatic force acting on a charge q is given by
where E is the electric field's intensity.
In our problem, the particle is an electron, so its charge is
. We know the intensity of the force, so we can find the magnitude of the electric field at the point where the electron is located:
where the negative sign means that the force and the electric field have opposite direction, because the charge is negative.
Answer:
Explain. yes it is possible because when the net force is zero and acceleration is zero. ... no, Neglecting air friction, the only force acting on this object is the force of gravity downward. There is no force of equal magnitude as the force of gravity acting in the opposite direction.
Given :
One particle, of mass m , moves with a speed v in the x-direction, and another particle, of mass 2 m , moves with a speed v/2 in the y-direction.
To Find :
The velocity of the center of mass of these two particles.
Solution :
Speed of mass m, .
Speed of mass 2m , .
Speed of center of mass is given by :
Hence, this is the required solution.
Answer:
1.96 J
Explanation:
From the law of conservation of energy
ΔU + ΔK = 0 where ΔU = internal energy change and ΔK = kinetic energy change. We neglect potential energy change since we are not given any information about it.
ΔU = -ΔK
ΔK = K₂ - K₁ where K₁ = initial kinetic energy and K₂ = final kinetic energy = 0 where ΔU = 0.382K₁
= 0.382mv²/2 where m = mass of spike = 0.71 kg and v = initial speed of spike = 3.8 m/s
= 0.382 × 0.71 kg × (3.8 m/s)²/2
= 1.96 J