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mixer [17]
3 years ago
7

If the standard kilogram bar kept in Paris were subjected to a net force of 1 newton, what acceleration would it have as a resul

t?
Physics
2 answers:
sdas [7]3 years ago
7 0

Answer:

It would have an acceleration of 1\frac{m}{s^{2}}

Explanation:

Let's start defining the newton unit (Newton : N)

The N definition is

N=Kg.(\frac{m}{s^{2}})

Where Kg is kilogram and \frac{m}{s^{2}} is an acceleration unit.

We can think the definition of newton as :

Given an object with a mass of 1 Kg subjected to a force of 1 N the acceleration that it experiments is 1\frac{m}{s^{2}}

As a result, the standard  kilogram bar kept in Paris will experiment an acceleration of 1\frac{m}{s^{2}} If we subject it to a net force of 1 N.

Klio2033 [76]3 years ago
3 0
1 m/s^2
Because of
F=ma
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3 years ago
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Peak wavelength of light coming from a star with a temp of 4300k
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During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying
Jlenok [28]

Well, I guess you can come close, but you can't tell exactly.

It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.

That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.

  98.5 m/s  =  √ [ (horizontal component)² + (vertical component)² ].

The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:

             Height = (1/2) (acceleration) (time²) .

If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.

                  Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
                 Height = (1/2) (9.81) (10.04)²

                            =   (4.905 m/s²) x (100.8 sec²)  =  494.43 meters.

As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.

If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component.  That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.   

8 0
3 years ago
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