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Sunny_sXe [5.5K]
2 years ago
12

Arrange the following decisions accordingly. Be guided by the chronological circumstances happened from the start of the game. U

se Roman Numerals in answering. 11. Hinto! Hiwalay! Hatol. Unang puntos, Bughaw! 12. Hinto! Pula, pangalawang laglag. Panalo, Bughaw! 13. Hinto! Hiwalay! Bughaw, 1 puntos. Unang paglabag! 14. Hinto! Bughaw, pangalawang paglabag! 15. Hinto! Pula, babala! 16. Hinto! Pula, unang laglag. 17. Hinto! Hiwalay! Hatol. Pula, I puntos. Bughaw, pangatlong paglabag! Panalo, Pula! 18. Hinto! Hiwalay! Bughaw, mag-ayos.​
Physics
1 answer:
Snezhnost [94]2 years ago
3 0

Answer: Use Roman Numerals in answering. 11. - 27310189. ... Hiwalay! Hatol. Unang puntos, Bughaw! 12. Hinto! Pula, pangalawang laglag. Panalo ...

Explanation:. Hinto! Hiwalay! Bughaw, 1 puntos. Unang paglabag! 14. Hinto! Bughaw, pangalawang paglabag

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What is one advantage of using electromagnets instead of permanent magnets
zimovet [89]
The main advantage of an electromagnet over a permanent magnet is that the magnetic field can be quickly changed by controlling the amount of electric current in the winding. However, unlike a permanent magnet that needs no power, an electromagnet requires a continuous supply of current to maintain the magnetic field.
5 0
3 years ago
Read 2 more answers
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
3 years ago
Can all rocks be weathered and eroded?how do you know?
skelet666 [1.2K]

Answer:yes

Explanation: Rock can be weathered by ice, acids, animals, temp and etc because the breaking down or dissolving of rock and minerals on the earths surface.

rocks can be eroded by wind, water, ice and, gravity be cause of an action, and friction.

(I think going off of what i learned.)

8 0
3 years ago
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A -0.06 charge that moves downward is in a uniform electric field with a strengthened of 200 N/C. What is the magnitude and dire
Nookie1986 [14]

Answer:

The magnitude of the force is 12 N Upwards

Explanation:

The force on a positive charge will be in the same direction as the field, but the force on a negative charge will be in the opposite direction to the field. Thus the direction of the force is upward.

Given;

magnitude of charge, q = 0.06 C

magnitude of electric field, E = 200 N/C

The magnitude of the force is given by;

F = qE

F = 0.06 x 200 N/C

F = 12 N Upwards

Therefore, the magnitude of the force is 12 N Upwards

5 0
3 years ago
During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa
maw [93]

Answer:

Explanation:

Given

Initial speed u=60\ mi/hr\approx 88\ ft/s

distance traveled before coming to rest d_1=120\ ft

using equation of motion

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

0-(88)^2=2\times a\times 120---1

for u_2=80\ mi/hr\approx 117.33\ ft/s

using same relation we get

0-(117.33)^2=2\times a\times (d_2)----2

divide 1 and 2 we get

(\frac{88}{117.33})^2=\frac{120}{d_2}

d_2=213.32\ ft

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0
3 years ago
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