Astone is catapulted at time t = 0, with an initial velocity of magnitude 18.0 m/s and at an angle of 45.0° above the horizontal. (neglect air resistance.) find its horizontal and vertical displacements from the catapult site at the following times after launch. (assume that the +x-axis is to the right and the +y-axis is up along the page.)

R = 33.1 m  and y = 8.27 m

Explanation:

This exercise is projectile launch, let's use the scope equation to find the displacement on the X axis

      R = vo2 sin 2θ / g

      R = 18 2 sin (2 45) /9.8

      R = 33.1 m

First let's look for the velocity components

       vox = vo cos θ

       voy = vo sin θ

       vox = 18 cos 45

       vox = 12.73 m / s

       voy = 18 sin 45

       voy = 12.73 m / s

To find the maximum vertical displacement, where the vertical velocity must be zero, we can use the equation

        ² = ² - 2 g y

         0 = ² - 2 g y

         y =² / 2g

          y = 12.73² / (2 9.8)

          y = 8.27 m

a=d/t

a="triangle"v/t

explanation:

a = acceleration

the answer is the first electromagnet (a). this is because it has more coils around the ferromagnetic material. increasing the number of coils increases the surface area in which the metal interacts with the current in the coil. this enables more atoms in the metal to become aligned hence becoming more magnetized.