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LenKa [72]
3 years ago
13

During a move, Jonas and Matías carry a 115kg safe to the third floor of a building, covering a height of 6.6m.

Physics
2 answers:
neonofarm [45]3 years ago
4 0

Answer:

work is =7590joules

power = 23watts

Neporo4naja [7]3 years ago
3 0

Answer:

1) 7590 Joules

2) 23 Watts

Explanation:

1) Work = force × distance

W = mgh

W = (115 kg) (10 m/s²) (6.6 m)

W = 7590 J

2) Power = work / time

P = W / t

P = (7590 J) / (330 s)

P = 23 W

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Mashutka [201]

Answer:

a) acceleration

Explanation:

Acceleration is, by definition, the change of an object's velocity.

4 0
2 years ago
8 POINTS EASY MULTIPLE CHOICE, BRAINLIST FOR CORRECT ANSWER:
zubka84 [21]
C) Increasing voltage
8 0
3 years ago
Hey everyone<br><br><br>what is optics??​
lana66690 [7]

Explanation:

Optics is a branch of physics that is the study of light and vision. ... The branch of physics dealing with the nature and properties of electromagnetic energy in the light spectrum and the phenomena of vision. In the broadest sense, optics deals with infrared light, visible light, and ultraviolet light.

7 0
3 years ago
Objects 1 and 2 attract each other with a gravitational force of 179 units. If the distance separating objects 1 and 2 is change
yaroslaw [1]

Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant throughout the universe.

D is the distance between the 2 objects.

the distance is now quadrupled.

Fgravitynew = G*(mass1*mass2)/(4D)² =

= G*(mass1*mass2)/(16D²) =

= (G*(mass1*mass2)/D²) / 16 = Fgravity/16

the new gravitational force will be 179/16 = 11.1875 units

3 0
2 years ago
Please help me
qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.

Now, the position function may look big and ugly in the picture.  But with the
exception of  't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation.  The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

Second derivative . . . y'' (t) = (a₀ - g) -  5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) -  (a₀ /t₀⁴ ) t⁴

and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

a₀ is the acceleration due to the model rocket engine thrust
     combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀  is how long the model rocket engine burns

Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.

The big name in model rocketry is Estes.  Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.


6 0
3 years ago
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