Equal to the angle of reflection.
Answer:
12mph in 2hrs and 3mph in 0.5hrs the total distance would be 12*2 and 3*0.5 which would be 24 and 1.5 so we add those 24+1.5= 25.5. The answer would be 25.5
Answer:
24,000 m
Explanation:
First find the rocket's final position and velocity during the first phase in the y direction.
Given:
v₀ = 75 sin 53° m/s
t = 25 s
a = 25 sin 53° m/s²
Find: Δy and v
Δy = v₀ t + ½ at²
Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²
Δy = 7736.8 m
v = at + v₀
v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)
v = 559.0 m/s
Next, find the final position of the rocket during the second phase (as a projectile).
Given:
v₀ = 559.0 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 15945.5 m
The total displacement is:
7736.8 m + 15945.5 m
23682.2 m
Rounded to two significant figures, the maximum altitude reached is 24,000 m.
Answer:
B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.
Explanation:
Impulse created by ball Y on ball X = 40 x 1/6 Ns
Ball X will also create impulse 40 / 6 on ball Y .
impulse = change in momentum .
impulse in Y = change in momentum in Y .
Initial momentum of Y = .5 x 5 = 2.5
Let final velocity of Y after collision be v in opposite direction .
change in momentum of Y = v - (-2.5 )
so,
v + 2.5 = 40 / 6 = 6.67
v = 4.17 m / s .
Option B is correct .
B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.
