Answer:
a) 0.618 ft/s
b) 3.04 ft/s
Explanation:
<u>Givens:</u>
Weight of swimmer A = 190 Ib.
Weight of swimmer B = 125 Ib.
Weight of the raft = 300 Ib.
Swimmer A walks toward swimmer B relative to the raft with a speed
= 2 ft/s
<em>a)</em><em> Conservation of linear momentum </em>
Since swimmer B does not move
Substitute from (2) and (3) into (1)
b) if the raft not to move
from (2)
substitute in (1)
Answer:
-1m/s
Explanation:
We can calculate the speed of block A after collision
According to collision theory:
MaVa+MbVb = MaVa+MbVb (after collision)
Substitute the given values
5(3)+10(0) = 5Va+10(2)
15+0 = 5Va + 20
5Va = 15-20
5Va = -5
Va = -5/5
Va = -1m/s
Hence the velocity of ball A after collision is -1m/s
Note that the velocity of block B is zero before collision since it is stationary
First, you set 58= 3.14 (pi) r^2. Solving for r you get about 4.3ft. You then use this radius in the circumference equation, C=2 3.14(pi) r. 2x3.14x4.3=27.004ft. I rounded a few times along the way so your answer is B.
9.8 is the gravitational field strength on earth, perhaps re-read the question