Answer:
The force per unit length (N/m) on the top wire is 16.842 N/m
Explanation:
Given;
distance between the two parallel wire, d = 38 cm = 0.38 m
current in the first wire, I₁ = 4.0 kA
current in the second wire, I₂ = 8.0 kA
Force per unit length, between two parallel wires is given as;
where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
Substitute the given values in the above equation and calculate the force per unit length
Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m
Answer:
Explanation:
Given that, the range covered by the sphere, , when released by the robot from the height,
, with the horizontal speed
is
as shown in the figure.
The initial velocity in the vertical direction is .
Let be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e.
will remain constant throughout the projectile motion.
So, if the time of flight is , then
Now, from the equation of motion
Where is the displacement is the direction of force,
is the initial velocity,
is the constant acceleration and
is time.
Here, and
(negative sign is for taking the sigh convention positive in
direction as shown in the figure.)
So, from equation (ii),
Similarly, for the launched height , the new time of flight,
, is
From equation (iii), we have
Now, the spheres may be launched at speed or
.
Let, the distance covered in the direction be
for
and
for
, we have
[from equation (iv)]
[from equation (i)]
(approximately)
This is in the points range as given in the figure.
Similarly,
[from equation (iv)]
[from equation (i)]
(approximately)
This is out of range, so there is no point for .
Hence, students must choose the speed to launch the sphere to get the maximum number of points.
