The line that should be secured first in pushing the boat
away from the dock in preparation to dock is the bow line. When the bow line is
secured, it is best to reverse it and turn to the dock, this will engage the
line to tighten in a way that will help it swing back in the dock.
Explanation:
Orbital speed= 2pi x radius / time period
=2pi x 1.5x10^11 / 365.25
=2.58x10^9m/day
I think it's a) 1st Newton's law... so sorry if it's wrong...
Answer:
The momentum is 1.94 kg m/s.
Explanation:
To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.
The potential energy 
of the compressed spring is given by
,
where 
is the length of compression and 
is the spring constant.
And the kinetic energy of the ball is
When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,
solving for 
we get:
And since momentum of the ball is 
,
Putting in numbers we get:
<span>net work = change in kinetic energy
for Block B, we just have the force from block A acting on it
F(ab)d= .5(1)vf² - .5(1)(2²)
F(ab)d= .5vf² - 2
Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction
10-F(ba)d = .5(4)vf² - .5(4)(2²)
10-F(ba)d = 2vf² - 8
F(ba)d = 18 - 2vf²
now we have two equations:
F(ba)d = 18 - 2vf²
F(ab)d= .5vf² - 2
since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab)
10-.5vf² + 2 = 2vf² - 8
12 - .5vf² = 2vf² - 8
20 = 2.5vf²
vf² = 8
they both will have the same velocity
KE of block A= .5(4)(2.828²) = 16 J
KE of block B=.5(1)(2.828²) = 4 J</span>
