Answer:
10.4 m/s
Explanation:
First, find the time it takes for the projectile to fall 6 m.
Given:
y₀ = 6 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.11 s
Now find the horizontal position of the target after that time:
Given:
x₀ = 6 m
v₀ = 5 m/s
a = 0 m/s²
t = 1.11 s
Find: x
x = x₀ + v₀ t + ½ at²
x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²
x = 11.5 m
Finally, find the launch velocity needed to travel that distance in that time.
Given:
x₀ = 0 m
x = 11.5 m
t = 1.11 s
a = 0 m/s²
Find: v₀
(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²
v₀ = 10.4 m/s
Answer:
The magnitude of the average angular acceleration of the disk is 
.
Explanation:
Given that,
Angular velocity, 
The disk comes to rest, 
Time, t = 0.234 s
We need to find the magnitude of the average angular acceleration of the disk. It is given by change in angular velocity per unit time. So,
So, the magnitude of the average angular acceleration of the disk is .
The best answer to fill in the blank should be ''continuation'' because series means a lot and lines are long, narrow, figures, so its continuation since its constant.
