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 .
Velocity is defined as Distance divided by Time.
In other words, V = D/T.
Now that we have our formula, we can solve.
Let's plug in the numbers we have.
We have 12m [East (direction not necessary when solving yet)] for our distance, and 0.15s as our time.
Divide the distance (12 /) by the time (0.15)
12 / 0.15 = 80.
Your velocity is 80 m/s [E]
I hope this helps!
Heat transfer is affected by color.
In the context of the loop and junction rules for electrical circuits, a junction is where three or more wires are joined.
Answer: Option 2
<u>Explanation:
</u>
An electrical circuits consists of many points like branch, loop, junction, series, bridge, etc. So, loops are the ones where the output of one circuit will act as feedback of the same circuit. If two or more wires passes through a single point, then that point is termed as junction.
If two or three junction connect each other they are termed as branch. Like these several other parameters are there with different rules in the circuit system. For electrical circuits, junction and loop rules state that a junction is the point where more wires joined together.
To solve this problem we will use the concepts related to hydrostatic pressure. Which determines the pressure of a body at a given depth of a liquid.
Mathematically this can be described as
Here
= Density
g = Gravity
h = Height (Depth)
If we replace the values given in the equation we will have to
Therefore the pressure at the bottom will be 9.8kPa