Equatorial currents are primarily westward. This is because the dominant current in the northern hemisphere has a clockwise direction, while the southern hemisphere has a counterclockwise direction. When these two currents meet at the equator, a common westward current exists.
<u>Answer:</u>
The final velocity of the two railroad cars is 1.09 m/s
<u>Explanation:</u>
Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by
where
V= Final velocity
M1= mass of the first object in kgs = 12000
M2= mas of the second object in kgs = 10000
V1= initial velocity of the first object in m/s = 2m/s
V2= initial velocity of the second object in m/s = 0 (given at rest)
Substituting the given values in the formula we get
V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s
Which is the final velocity of the two railroad cars
Use kinematic equations to solve:
1) yf = yo + vo*t + 1/2at²
yf = final height
yo = initial height
vo = initial velocity
a = acceleration
t = time
yf - yo = vo*t + 1/2at²
yf - yo = h
vo = 0
Thus,
h = 1/2at²
h = 1/2(9.8)(12)² = 705.6 m
2) vf = vo + at
vo = 0
Thus,
vf = at
vf = (9.8)(12) = 117.6 m/s
Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.
If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.
If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is
y(t) = y₀ + M sin(2π t/15) .
The vertical speed of the deck is y'(t) = M (2π/15) cos(2π t/15)
and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)
= - (4 π² M / 15²) sin(2π t/15)
= - 0.1755 M sin(2π t/15) .
There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.
The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).
At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.
Now I'm going to wave my hands in the air a bit:
Apparent weight = (apparent mass) x (real acceleration of gravity)
(Apparent mass) = (65/60) = 1.08333 x real mass.
Apparent 'gravity' = 1.08333 x real acceleration of gravity.
The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.
0.08333 G = 0.1755 M
The 'M' is what we need to find.
Divide each side by 0.1755 : M = (0.08333 / 0.1755) G
'G' = 9.0 m/s²
M = (0.08333 / 0.1755) (9.8) = 4.65 meters .
That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .
Uniform velocity means no Net force and therefore no acceleration. Acceleration only happens when the velocity changes.
