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
That's electric "current", usually described in 'Amperes'.
1 Ampere = 1 Coulomb per second ... "the rate at which
charge passes a point in the circuit".
Answer:
The distance of the first bright fringe is given as 
The distance of the second dark fringe from the central bright fringe is given as 
Explanation:
From the question we are told that
The slit separation distance is 
The distance of the slit from the screen is 
The wavelength is 
For constructive interference to occur the distance between the two slit is mathematically represented as

Where m is the order of the fringe which has a value of 1 for first bright fringe
Substituting values


For destructive interference to occur the distance between the two slit is mathematically represented as
![Y_D = [n + \frac{1}{2} ] \frac{\lambda D}{d}](https://tex.z-dn.net/?f=Y_D%20%20%3D%20%20%5Bn%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5D%20%5Cfrac%7B%5Clambda%20D%7D%7Bd%7D)
m = 2
so the formula to get the dark fringe is 
Now substituting values
![Y_D = [ 1 + \frac{1}{2} ] * \frac{633 *10^{-9} * 3.23 }{0.00115}](https://tex.z-dn.net/?f=Y_D%20%3D%20%5B%201%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5D%20%2A%20%5Cfrac%7B633%20%2A10%5E%7B-9%7D%20%2A%203.23%20%7D%7B0.00115%7D)


Answer:
E = 0.0130 V/m.
Explanation:
The electric field is related to the potential difference as follows:

<u>Where:</u>
E: is electric field
ΔV: is the potential difference = 3.95 mV
d: is the distance of a person's chest = 0.305 m
Then, the electric field is:

Therefore, the maximum electric field created is 0.0130 V/m.
I hope it helps you!