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:
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Answer:
0.231 N
Explanation:
To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

So the force acting on the other end to generate this torque mush be:

Answer:
D.
Explanation:
A solar system is a collection of planets, their moons, and other objects in orbit around a central star.
the friction forces are smaller than the forward force