Answer:
Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from
The answer to the question is
The speed 
of the electron when it is 10.0 cm from charge Q₁
= 7.53×10⁶ m/s
Explanation:
To solve the question we have
Q₁ = 3.45 nC = 3.45 × 10⁻⁹C
Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C
2·d = 50.0 cm
a = 10.0 cm
q = -1.6×10⁻¹⁹C
Also initial kinetic energy = 0 and
Initial electric potential energy = 
Final kinetic energy due to motion = 0.5·m·v²
Final electric potential energy = 
From the energy conservation principle we have
Solving for v gives
where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg
gives v =7528188.32769 m/s or 7.53×10⁶ m/s
= 7.53×10⁶ m/s
Answer:
a
b
Explanation:
From the question we are told that
The speed of the airplane is 
The angle is 
The altitude of the plane is 
Generally the y-component of the airplanes velocity is
=> 
=> 
Generally the displacement traveled by the package in the vertical direction is
=> 
Here the negative sign for the distance show that the direction is along the negative y-axis
=> 
Solving this using quadratic formula we obtain that
Generally the x-component of the velocity is
=> 
=> 
Generally the distance travel in the horizontal direction is
=> 
=>
Generally the angle of the velocity vector relative to the ground is mathematically represented as
![\beta = tan ^{-1}[\frac{v_y}{v_x } ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%20tan%20%5E%7B-1%7D%5B%5Cfrac%7Bv_y%7D%7Bv_x%20%7D%20%5D)
Here 
is the final velocity of the package along the vertical axis and this is mathematically represented as
=> 
=> 
and v_x is the final velocity of the package which is equivalent to the initial velocity 
So
![\beta = tan ^{-1}[-130.05}{57.96 } ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%20tan%20%5E%7B-1%7D%5B-130.05%7D%7B57.96%20%7D%20%5D)
The negative direction show that it is moving towards the south east direction
