Solving this using the time, we know that range = horizontal velocity x time of flight
since
there are no horizontal forces acting on the ball, there are no
horizontal accelerations and the initial horizontal velocity of 36 cos
28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have
range = 36 cos 28 x 3.44 s = 109.3 m
Answer:
The magnitude of the force between the two parallel wires is 0.0111 N.
Explanation:
Given;
length of the two parallel wires, L = 42 m
distance between the two wires, r = 0.03 m
current in both wires, I₁, I₂ = 6.3 A
Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.
Answer:

Explanation:
As we know by radioactivity law

so here we will have


now we will have


now we also know that



I think it’s 15cm
Might be 7cm