Answer:
Statement:
The electric current passing through a conductor is directly proportional to the potential difference across its ends provided temperature and other physical conditions remain constant.
Explanation:
Current is directly proportional to voltage loss through a resistor. That is, if the current doubles, then so does the voltage. To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R).
V∝I or I∝V⇒V=IR.
Answer:
The direction of the displacement is in North-West.
Explanation:
Resultant displacement D is
Here the direction is
Then the direction is 
North-west.
Answer:
work = 1125 [J]
Explanation:
To solve this problem we must remember the definition of power, which is defined as the relationship between work and time. The power can be calculated using the following equation:
Power = work/time
Power = 12.5 [w]
work = joules [J]
time = 1.5 [min] = 90 [s]
work = 12.5*90
work = 1125 [J]
First, find the amount of time for the dart to hit the board using this equation: t = d/v
t = 2 m/ 15 m/s = 0.133 s
Then, find the height the dart has fallen from its initial point using this equation: h = 0.5gt²
h = 0.5(9.81 m/s²)(0.133 s)² = 0.0872 m or 8.72 cm
Since the diameter of the bull's eye is only 5 cm, and you started at the same level of the top of the bull's eye, that means the maximum allowance would only be 5 cm. Since it exceeded to 8.72 cm, it means that <em>Veronica will not hit the bull's eye.</em>
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 .
