Answer:

Explanation:
Given,
mass of the object,m = 0.30 Kg
initial speed, v_i = 3 m/s
time of collision = 0.20 s
final speed, v_f = -4 m/s
Impulse = change in momentum



Hence, impulse of the object is equal to 
Answer:

Explanation:
The electrostatic potential energy for pair of charge is given by
U=1/4π∈₀×(q₁q₂/r)
Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs or charges.For three equal charges on the corners of an equilateral triangle,the electrostatic potential energy is given by:
U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)
U=3×1/4π∈₀×(q²/r)
Substitute given values
So
Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m
Answer:
3.88 * 10^(-15) J
Explanation:
We know that the Potential energy of the electron at the beginning of its motion is equal to the Kinetic energy at the end of its motion, when it reaches the plates.
First, we get the potential and potential energy:
Electric potential = E * r
E = electric field
r = distance between plates
Potential = 2.2 * 10^6 * 0.011
= 2.42 * 10^4 V
The relationship between electric potential and potential energy is:
P. E. = q*V
q = charge of electron = 1.602 * 10^(-19) C
P. E. = 2.42 * 10^4 * 1.602 * 10^(-19)
P. E. = 3.88 * 10^(-15) J