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Thepotemich [5.8K]
3 years ago
13

1.Sobre a queda-livre,assinale V para verdadeiro e F para falso

Physics
1 answer:
lisov135 [29]3 years ago
8 0

Answer:pelo o que eu sei é ..

V

V

V

F

F

F

Explanation:

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What is 178.57 rounded to the nearest whole number
postnew [5]
<h2><em>178.57</em></h2>

<h2><em>=</em><em> </em><em>178</em><em>.</em><em>6</em></h2>

<h2><em>=</em><em> </em><em>179</em></h2>

<h2><em>THANK</em><em> </em><em>YOU</em><em> </em></h2>
4 0
3 years ago
Read 2 more answers
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction
Sophie [7]

Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

         F - fr = 0

         F = fr

         fr = 52.1 N

Now we can work  with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

Y axis

        N - W₁ -W₂ = 0

        N = W₁ + W₂

as the two blocks are identical

        N = 2W

X axis

        F - fr₁ - fr₂ = 0

        F = fr₁ + fr₂

indicates that the lower block is moving below block 1, therefore the upper friction force is

          fr₁ = 52.1 N

          fr₁ = μ N

a

s the normal in the lower block of twice the friction force is

          fr₂ = μ 2N

          fr₂ = 2 μ N

          fr₂ = 2 fr₁

we substitute

          F = fr₁ + 2 fr₁

          F = 3 fr₁

          F = 3  52.1

          F = 156.3 N

7 0
3 years ago
The universal law of gravitation states that the force of attraction between two objects depends on which quantities?
Ronch [10]

Answer:

D. the masses of the objects and the distance between them

Explanation:

Gravitation is a force, a force doesn't care about the shape or density of objects, only about their masses... and distances.

And you can get it using the following equation:

f = \frac{Gm_{1}m_{2} }{d^{2} }

Where :

G is the universal gravitational constant : G = 6.6726 x 10-11N-m2/kg2

m represent the mass of each of the two objects

d is the distance between the centers of the objects.

4 0
3 years ago
A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depend
liq [111]
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

                     (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere. 
Only the initial and final y-coordinates matter.

We know that    F = - 2.85 y².  (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From  y=0  to  y=2.40  is a distance of  2.40  upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .


Now, integral of (y² dy)  =  1/3  y³ .

Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the  -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40.  Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
by some other route.  WHATEVER the route is, the work done by ' F ' 
is going to total up to be  -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then.  But that's my answer
and I'm stickin to it.  If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
3 0
3 years ago
How to find resultant force?
LuckyWell [14K]

Explanation:

To find the resultant force subtract the magnitude of the smaller force from the magnitude of the larger force. The direction of the resultant force is in the same direction as the larger force

4 0
3 years ago
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