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aleksklad [387]
3 years ago
10

Can someone help me please

Physics
1 answer:
TEA [102]3 years ago
8 0

Answer:

D

Explanation:

Because it is impossible for it to show the real depth of the ocean and how deep it is

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A string attached to an airborne kite was maintained at an angle of 40.0 with the ground. if 120m of string was reeled in to ret
AleksAgata [21]
The hypotenuse is measured at 120 meters of string, and you need to solve for the leg of the triangle that is horizontal. The degree is 40, so use trigonometry to figure it out. 
Cosin (40) is equal to around .766 
Adjacent/Hypotenuse 
x/120 = cos40 
Answer: 91.92533. 
If you use 3 significant figures it should be 91.9 meters.
5 0
3 years ago
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6 0
2 years ago
If you travel 1.7 km north from your house at noon, and at 6:00 PM you travel 5.4 km south, what is your displacement? 3.7 km no
Yakvenalex [24]

<u>Answer</u>

3.7 Km south


<u>Explanation</u>

The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.

Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).

Now lets add the two values.

(+1.7) + (-5.4) = 1.7 - 5.4

                    = - 3.7 Km      But negative was towards south.

∴ Answer = 3.7 Km south.


6 0
3 years ago
A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

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 .

8 0
2 years ago
Same diagram... Which location shows SUMMER in the SOUTHERN hemisphere? *
LiRa [457]
Location 1 is correct
7 0
3 years ago
Read 2 more answers
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