(10 points) A spring with a 7-kg mass and a damping constant 12 can be held stretched 1 meters beyond its natural length by a force of 4 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c1eαt+c2eβt where α= (the larger of the two) β=

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Answer 1
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Related Questions

You're carrying a 3.6-m-long, 21 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. Part A Part complete How much force must you exert to keep the pole motionless in a horizontal position? Express your answer in newtons. F = 114 N Previous Answers
Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 2 seconds.
As you take the stoppered part of the tube up the staircase you begin to see the water level drop around the 4th floor. As you continue up it does not continue up with you but stays at a constant level. What does that mean?a. The pressure in the tubing is equal to the barometric pressure.b. The tubing was unable to supply any more water to the tube for use.c. The pressure outside the tube is higher that the water pressure inside the tube.
A block m1 rests on a surface. A second block m2 sits on top of the first block. A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.(a)What is the normal force exerted by the surface on the bottom block? (Use the following as necessary: m and g as necessary.)
Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 6.7Ω and 15.9 W, 30.4Ω and 9.12 W, and 16.3Ω and 12.3 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?

We know that there is a relationship between work and mechanical energy change. Whenever work is done upon an object by an external force (or non-conservative force), there will be a change in the total mechanical energy of the object. If only internal forces are doing work then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. Think of a real-life situation where we make use of this conservation of mechanical energy (where we can neglect external forces for the most part). Describe your example and speak to both the kinetic and potential energy of the motion.

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Answer:

* roller skates and ice skates.

* roller coaster

Explanation:

One of the best examples for this situation is when we are skating, in the initial part we must create work with a force, it compensates to move, after this the external force stops working and we continue movements with kinetic energy, if there are some ramps, we can going up, where the kinetic energy is transformed into potential energy and when going down again it is transformed into kinetic energy. This is true for both roller skates and ice skates.

Another example is the roller coaster, in this case the motor creates work to increase the energy of the car by raising it, when it reaches the top the motor is disconnected, and all the movement is carried out with changes in kinetic and potential energy. In the upper part the energy is almost all potential, it only has the kinetic energy necessary to continue the movement and in the lower part it is all kinetic; At the end of the tour, the brakes are applied that bring about the non-conservative forces that decrease the mechanical energy, transforming it into heat.

A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initialspeed of 32 m/s, what is its new speed?
A. 17 m/s
B. 15 m/s
C. 47 m/s
D. 32 m/s

Answers

Answer:

17 m/s

Explanation:

Using formula a = (v-u) /t

acceleration a =  -1.5 m/s2

final velocity v = unknown

initial velocity u = 32 m/s

time t = 10s

-1.5 = (v-32)/10

-15 = v - 32

-15 + 32 = v

v = 17 m/s

How does clothing and body language affect a job interview?

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Answer:

please give me brainlist and follow

Explanation:

Whilst first impressions can be important, your body language during the interview can make or break your overall performance. With experts saying that between 75-90% of communication is non-verbal, it is important to think about what your body is saying about you during an interview.

For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8 cosθ) m, where t is measured in seconds, Determine the magnitudes of the car's velocity and acceleration when t=4s .

Answers

Answer:

Velocity = v = 2.24m/s

Acceleration = a = 0.20m/s²

Explanation:

Please see attachment below.

Given

z=(−8 cosθ) and θ = 0.3t

z = -8Cos (0.3t)

V = dz/dt

a = v²/R.

Please see full solution below.

Final answer:

The roller coaster's velocity and acceleration at t=4 seconds is 7.64 m/s and 0.57 m/s² respectively.

Explanation:

The question is about understanding kinematics in cylindrical coordinates to analyze the motion of a roller coaster car. First, we need to understand that in polar coordinates, θ is changing with time t. So, the velocity vector v will have two components, one in the θ direction (rθ') and another in the z direction (z'). Given θ = 0.3t, we differentiate θ with respect to time to get θ' or dθ/dt = 0.3 rad/sec. Then, the z component of the velocity can be calculated by differentiating the equation of motion in the z-direction, z = -8 cos(θ), with respect to time. This gives z' = 8(0.3)sin(0.3t). So, at t=4s, z' = 8(0.3)sin(1.2) = 1.89 m/s. We then calculate rθ' = r*dθ/dt = 25*0.3 = 7.5 m/s.

The magnitude of velocity can then be calculated using the Pythagorean theorem: √((rθ')² + (z')²) = √((7.5)² + (1.89)²) = 7.64 m/s .

In a similar way, we can find the acceleration components. Given that r=25 m and is constant, radial acceleration is zero ( ar = r*(θ')²). The tangential acceleration is at = r*θ'' = r*d²θ/dt² =0 m/s² and z'' = dz'/dt = 8*0.3²*cos(0.3t). So, at t = 4s, z'' = 8(0.09)cos(1.2) = 0.57 m/s². The magnitude of the acceleration is given by √((ar)² + (at)² +(z'')²) = √((0)² + (0)² +(0.57)²)= 0.57 m/s².

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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)

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High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of the track depends on the speed of the particle and its lifetime. A particle moving at 0.993c leaves a track 1.15 mm long. What is the proper lifetime of the particle

Answers

Answer:

Lifetime = 4.928 x 10^-32 s  

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s  

Final answer:

To find the proper lifetime of the particle, we can use the time dilation equation and the Lorentz factor. Plugging in the given values, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

Explanation:

To find the proper lifetime of the particle, we can use the time dilation equation, which states that the proper time (time experienced in the frame of reference of the particle) is equal to the time observed in the laboratory frame of reference divided by the Lorentz factor. The Lorentz factor can be calculated using the equation γ = 1/√(1 - (v/c)^2), where v is the velocity of the particle and c is the speed of light. Given that the particle is moving at 0.993c, the Lorentz factor is approximately 22.82.

Next, we can use the equation Δx = βγcτ, where Δx is the length of the track, β is the velocity of the particle in units of the speed of light (v/c), γ is the Lorentz factor, c is the speed of light, and τ is the proper lifetime of the particle. Plugging in the given values, we have 1.15 mm = 0.993 * 22.82 * c * τ. Solving for τ, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

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