A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top of a relaxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After striking the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact

Answers

Answer 1
Answer:

Answer:

y_(max) = 0.829\,m

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

v = \sqrt{(0.8\,(m)/(s))^(2) + 2\cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m)}

v = 2.913\,(m)/(s)

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

(3\,kg)\cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m) + (1)/(2)\cdot (3\,kg)\cdot (2.913\,(m)/(s) )^(2) = (3\,kg) \cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s) ^(2)

After some algebraic handling, a second-order polynomial is formed:

12.728\,J = (1)/(2)\cdot (2000\,(N)/(m) )\cdot (\Delta s)^(2) - (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot \Delta s

1000\cdot (\Delta s)^(2)-29.421\cdot \Delta s - 12.728 = 0

The roots of the polynomial are, respectively:

\Delta s_(1) \approx 0.128\,m

\Delta s_(2) \approx -0.099\,m

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

\Delta s \approx 0.128\,m

The maximum height that the block reaches after rebound is:

(3\,kg) \cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s)^(2) = (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot y_(max)

y_(max) = 0.829\,m

Answer 2
Answer:

Answer:

0.81 m

Explanation:

In all moment, the total energy is constant:

Energy of sistem = kinetics energy + potencial energy = CONSTANT

So, it doesn't matter what happens when the block hit the spring, what matters are the (1) and (2) states:

(1): metal block to 0.8 m above the floor

(2): metal block above the floor, with zero velocity ( how high, is the X)

Then:

E_(kb1) + E_(gb1)  = E_(kb2) + E_(gs2)

E_(kb1) + E_(gb1)  = 0 + E_(gs2)

(1)/(2)*m*V_(b1) ^(2)   + m*g*H_(b1)  = m*g*H_(b2)

H_(b2)  =  (V_(b1) ^(2) )/(2g)  + H_(b1)

Replacing data:

H_(b2)  =  (0.44^(2) )/(2*9.81)  + 0.8

H_(b2)  =  (0.44^(2) )/(2*9.81)  + 0.4

HB2 ≈ 0.81 m


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Answers

Answer:

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Explanation:

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Explanation:

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G Water enters a house through a pipe 2.40 cm in diameter, at an absolute pressure of 4.10 atm. The pipe leading to the second-floor bathroom, 5.20 m above, is 1.20 cm in diameter. The flow speed at the inlet pipe is 4.75 m/s a) What is the algebraic expression for flow speed in the bathroom?
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Answers

Answer:

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B) V₂ = 19 m /s

C) P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

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Where;

A₁ and A₂ are area of cross section V₁ and V₂ are velocity of flow at two places along pipe.

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π x 1.2² x 4.75 = π x 0.6² x V₂

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V₂ = 19 m /s

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P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

Where;

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Answers

Answer:

Explained

Explanation:

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To see how two traveling waves of the same frequency create a standing wave. Consider a traveling wave described by the formula y1(x,t)=Asin(kx−ωt)
This function might represent the lateral displacement of a string, a local electric field, the position of the surface of a body of water, or any of a number of other physical manifestations of waves.
1. Find ye(x) and yt(t). Keep in mind that yt(t) should be a trigonometric function of unit amplitude.
2. At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)?
3. At certain times, the string will be perfectly straight. Find the first time t1>0 when this is true.
4. Which one of the following statements about the wave described in the problem introduction is correct?
A. The wave is traveling in the +x direction.
B. The wave is traveling in the −x direction.
C. The wave is oscillating but not traveling.
D. The wave is traveling but not oscillating.
Which of the expressions given is a mathematical expression for a wave of the same amplitude that is traveling in the opposite direction? At time t=0this new wave should have the same displacement as y1(x,t), the wave described in the problem introduction.
A. Acos(kx−ωt)
B. Acos(kx+ωt)
C. Asin(kx−ωt)
D. Asin(kx+ωt)

Answers

The definition of standing wave and trigonometry allows to find the results for the questions about the waves are:

      1. For the standing wave its parts are: spatial y_e = A' \ sin \ kx  and

         temporal part y_t = A' \ cos \ wt

      2. The string moves with an oscillating motion  y = A’ cos wt.

      3. Thefirst displacement is zero for  t = (\pi )/(2w)  

      4. the correct result is:

          A. The wave is traveling in the +x direction.

      5. The correct result is:

          D. Asin(kx+ωt)

Traveling waves are periodic movements of the media that transport energy, but not matter, the expression to describe it is:

       y₁ = A sin (kx -wt)

Where A is the amplitude of the wave k the wave vector, w the angular velocity and x the position and t the time.

1. Ask us to find the spatial and temporal part of the standing wave.

To form the standing wave, two waves must be added, the reflected wave is:

       y₂ = A sin (kx + wt)

The sum of a waves

       y = y₁ + y₂

       y = A (sin kx-wt + sin kx + wt)

We develop the sine function and add.

       Sin (a ± b) = sin a cos b ± sin b cos a

The result is:

       y = 2A sin kx cos wt

They ask that the function be unitary therefore

The amplitude  of each string

        A_ {chord} = A_ {standing wave} / 2

The spatial part is

          y_e= A 'sin kx

The temporary part is:

          y_t = A ’cos wt

2. At position x = 0, what is the displacement of the string?

          y = A ’cos wt

The string moves in an oscillating motion.

3. At what point the string is straight.

When the string is straight its displacement is zero x = 0, the position remains.

           y = A ’cos wt

For the amplitude of the chord to be zero, the cosine function must be zero.

           wt = (2n + 1) (\pi)/(2)  

the first zero occurs for n = 0

          wt = (\pi )/(2)  

           t = (\pi )/(2w)

4) The traveling wave described in the statement is traveling in the positive direction of the x axis, therefore the correct statement is:

         A. The wave is traveling in the +x direction.

5) The wave traveling in the opposite direction is

            y₂ = A sin (kx + wt)

The correct answer is:

            D.     Asin(kx+ωt)

In conclusion using the definition of standing wave and trigonometry we can find the results for the questions about the waves are:

     1. For the standing wave its parts are: spatial y_e = A' \ sin \ kx  and

         temporal part y_t = A' \ cos \ wt

      2. The string moves with an oscillating motion  y = A’ cos wt.

      3. Thefirst displacement is zero for  t = (\pi )/(2w)  

      4. the correct result is:

          A. The wave is traveling in the +x direction.

      5. The correct result is:

          D. Asin(kx+ωt)

Learn more about standing waves here:  brainly.com/question/1121886

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Answers

What is pseudo force?

A pseudo force, also called a fictitious force or an inertial force, is an apparent force that acts on all bodies whose motion is described using a non-inertial frame of reference, such as a rotating reference frame.