You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

Response:

210.3 degrees

Justification:

The total force acting on charge A is 59.5 N

Apply the x and y components of the net force to determine the direction

atan (y/x)

This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>

Answer:

x = v₀ cos θ   t,   y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This pertains to a projectile motion scenario. Here, we will express the equations for both the x and y dimensions.

Now, we will apply trigonometry to determine the initial velocity components.

              sin θ = / v₀

              cos θ = v₀ₓ / v₀

              v_{y} = v_{oy} sin θ

              v₀ₓ = v₀ cos θ

Next, let's formulate the equations of motion.

X axis

         x = v₀ₓ t

         x = v₀ cos θ   t

        vₓ = v₀ cos θ

Y axis

        y = y₀ + t - ½ g t2

        y = y₀ + v₀ sin θ t - ½ g t2

        v_{y} = v₀ - g t

       v_{y} = v₀  sin θ - gt

        = v_{oy}^2 sin² θ - 2 g y

It is evident that the major distinction lies in the fact that in an inclined launch compared to a horizontal one, the velocity comprises different components

To find the mass using a weight of 1.4 N:
1.4/9.8 = 0.1428 kg
The momentum is calculated as 0.1428 multiplied by 44.7, which is 6.38 kgm/s.