Answer:
The value is 
Explanation:
From the question we are told that
The distance of separation is 
The current on the one wire is 
The current on the second wire is 
Generally the magnitude of the field exerted between the current carrying wire is

Here
is the magnetic field due to the first wire which is mathematically represented as

Here
is the distance to the half way point of the separation and the value is

is the magnetic field due to the first wire which is mathematically represented as

Here
is the distance to the half way point of the separation and the value is
This means that 
So

=> 
=> 
=> 
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.
First, we determine how long the parcel will fall using:
s = ut + 1/2 at²
where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.
5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds
A
First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
A becuz its at da it dont got no wa
The answer is the less dense plate slides over the denser plate.