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Triss [41]
3 years ago
11

What type of wave is sound? a. longitudnal b. surface c. tranverse, or d.. light ? ?

Physics
2 answers:
AVprozaik [17]3 years ago
7 0

Answer:

Sound is a series of compression waves that moves through air or other materials. Sound waves are a longitudinal wave. The speed of sound is about 343 m/s which is much slower than the speed of light.

Explanation:

I hoped this helped

bagirrra123 [75]3 years ago
3 0
A longitudinal wave is a type of sound
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A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is
Damm [24]

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

        p₀ = m v

final attempt. after separation

       p_{f} = m /2  0 + m /2 v_{f}

       p₀ = p_{f}

       m v = m /2 v_{f}

       v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

     

initial energy

         K₀ = ½ m v²

final energy

        K_{f} = ½ m/2  0 + ½ m/2 v_{f}²

        K_{f} = ¼ m (2v)²

        K_{f} = m v²

         

the expression for work is

         W = ΔK = K_{f} - K₀

         W = m v² - ½ m v²

         W = ½ m v²

5 0
3 years ago
What is the similarity between pushing and lifting? What is the difference?
MA_775_DIABLO [31]

Answer:

the similarity is that hands are applied and force

7 0
2 years ago
High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines has a cross
goblinko [34]

Answer:

Explanation:

For resistance of a wire , the formula is as follows

R = ρ L / S

where ρ is specific resistance , L is length and S is cross sectional area

Given L = 14 000 m ,

S = 4.8 x 10⁻⁴ m²

specific resistance of aluminum = 2.8 x 10⁻⁸ ohm-meter

Putting the values in the formula

R = 2.8 x 10⁻⁸ x 14 x 10³ /  (4.8 x 10⁻⁴ )

R = 0.8167 ohm .

= .82 ohm .

5 0
2 years ago
A 4113 N piano is to be pushed up a(n) 3.99 mfrictionless plank that makes an angle of 25.5â—¦with the horizontal.Calculate the
Serhud [2]

Answer:

W = 14.8 kJ

Explanation:

W = F S cos ∅

W = 4113 x 3.99 x cos 25.5

W = 16410.87 x 0.9025 = 14810.8 J or 14.8 kJ

3 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
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