The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.
<h3>What is the law of conservation of momentum?</h3>
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
m is the mass =6.0 kg
t is the time interval=2 second
From Newton's second law;
From the graph;
The change in the momentum is;
Hence, the resulting change in momentum of the system will be +18.6 Ns.
To learn more about the law of conservation of momentum, refer;
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Explanation:
The formula to determine the eccentricity of an ellipse is the distance between foci divided by the length of the major axis
To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.
For the particular case of the Cube with equal sides its volume is determined by
In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say
In this way the net volume would be
We need to find the mass, but we have the Weight and Gravity so from Newton's second Law
PART A) From the relation of density as a unit of mass and volume we have to
PART B) To find the weight of the cube then we apply the ratio of
Answer:
speed of the charge electric is v = - (Eo q/m) cos t
Explanation:
The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,
F = q Eo sint
a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity
F = ma
q Eo sint = ma
a = Eo q / m sint
a = dv / dt
dv = adt
∫ dv = ∫ a dt
v-vo = I (Eoq / m) sin t dt
v- vo = Eo q / m (-cos t)
We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0
v = - (Eo q / m) cos t
b) Kinetic energy
K = ½ m v2
K = ½ m (Eoq / m)²2 (sint)²
K = ¹/₂ Eo² q² / m sin² t
c) The average kinetic energy over a period
K = ½ m v2
<v2> = (Eoq / m) 2 <cos2 t>
The average of cos2 t = ½, substitute and calculate
K = ½ m (Eoq / m)² ½
K = ¼ Eo² q² / m
