Answer: True
Explanation: When light is reflected off lets say a mirror it is bent and changes direction to bounce off of another wall or object. For example if you take a flash light and shine it into a mirror the light reflects into a different direction your welcome
<span> gravitational force varies based on 1/r^2
when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.
</span><span>As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)
</span>hope this helps
1. Answer: components
A two dimensional vector can be divided into two parts called horizontal component and vertical component.
A three dimensional vector can be divided into three components: one along x-axis, one along y-axis and one along z-axis.
Hence, the vector parts that add up to the resultant are called components.
2. Answer: 5 miles.
The resultant distance along the straight line from the starting point to the end point would be the displacement.
The displacement would be equal to the magnitude of the hypotenuse formed in the right triangle.
Displacement, 
3. Answer: Scalar
A scalar quantity has only magnitude. For example, speed and distance are scalar quantities and can be normally added to find the total.
A vector quantity has both magnitude as well as direction. The components are summed according to vector addition rules. For example, velocity, acceleration, force etc.
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
