Unscrambling
1. resting heart rate
2. overload
3. workout
4. specificity
5. cool-down
6. progression
7. warm-up
8. the last one can only be instance, but there was a typo on the paper.
Answer:
the energy of the spring at the start is 400 J.
Explanation:
Given;
mass of the box, m = 8.0 kg
final speed of the box, v = 10 m/s
Apply the principle of conservation of energy to determine the energy of the spring at the start;
Final Kinetic energy of the box = initial elastic potential energy of the spring
K.E = Ux
¹/₂mv² = Ux
¹/₂ x 8 x 10² = Ux
400 J = Ux
Therefore, the energy of the spring at the start is 400 J.
Answer:
Part A
Coriolis effect is used to describe how objects which are not fixed to the ground are deflected as they travel over long distances due to the rotation of the Earth relative to the 'linear' motion of the objects
Due to the Coriolis effect the wind flowing towards the Equator from high pressure belts in the subtropical regions in both the Northern and Southern Hemispheres are deflected towards the western direction because the Earth rotates on its axis towards the east
Part B
In the Northern Hemispheres, the winds are known as northeasterly trade winds and in the Southern Hemisphere, they are known as the southeasterly trade wind. Therefore, Coriolis effect has the same effect on the direction of the Trade Winds in the Southern Hemisphere as it does in the Northern Hemisphere
Explanation:
Answer:
Sedimentary rocks are types of rock that are formed by the deposition and subsequent cementation of that material at the Earth's surface and within bodies of water. Deposition means that all the sediments, soil, and rocks are all compressed (tightly pressed into each other) and create sedimentary rocks.
I hope this helps! :)
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 .
