Amissile is moving 1810 m/s at a 20.0 degree angle. it needs to hit a target 19,500 m away in a 32 degree direction in 9.20 seconds. what is the direction of the acceleration that the engine must produce?

79.1°

Explanation:

Given:

x₀ = 0 m

y₀ = 0 m

x = 19500 cos 32°

y = 19500 sin 32°

v₀x = 1810 cos 20°

v₀y = 1810 sin 20°

t = 9.20

Find:

ax, ay, θ

First, in the x direction:

x = x₀ + v₀ t + ½ at²

19500 cos 32° = 0 + (1810 cos 20°) (9.20) + ½ ax (9.20)²

16537 = 15648 + 42.32 ax

ax ≈ 21.01

And in the y direction:

y = y₀ + v₀ t + ½ at²

19500 sin 32° = 0 + (1810 sin 20°) (9.20) + ½ ay (9.20)²

10333 = 5695 + 42.32 ay

ay ≈ 109.6

The direction of the acceleration is therefore:

θ = atan(ay / ax)

θ = atan(109.6 / 21.01)

θ ≈ 79.1°

79.1°

Explanation:

A missile is moving 1810 m/s at a 20.0 degree angle. It needs to hit a target 19,500 m away in a 32 degree direction in 9.20 seconds. The direction of the acceleration that the engine must produce is 79.1°

it should be -8.6 cm hope this out

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge