The 55g arrow is launched so that it hits and embeds in a 3.54kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.49m higher than the block's starting point. How fast was the arrow moving before it joined the block? THE ANSWER FOR PEOPLE WITH MASTERING PHYSICS CUZ MOST OF THEM ARE WRONG ON PE = mgh = (3.54 + 0.55)kg * 9.8m/s² * 0.49m = 17.3 J

which means the post-collision KE = 17.3 J = ½ * 3.595kg * v², and

v = 3.10 m/s

Now conserve momentum: final Σ mv = initial Σ mv

3.595kg * 3.10m/s = 0.055kg * V

V = 203 m/s ≈ 200 m/s

answer: i don't know the answer good luck

explanation:

v 154.9= 0.9 length length s little over 154.9cm

i'm on my way to history not a mistory.

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.