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just olya [345]
2 years ago
5

A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 2

00 N/m.
The spring-block system is initially stretched a distance “d” beyond the spring's equilibrium position and is released from rest. As the block moves to the right, its speed is 50.0 cm/s when the spring-block system is 4.00 cm from the equilibrium position.
Find d, the initial displacement of the spring-block system (in cm).

Physics
1 answer:
Tom [10]2 years ago
4 0

the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:

  • The range of motion is: A = 4.44 cm

<h3>Oscillatory movement.</h3>

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

            x = A cos (wt + Ф)

            w² = k/m

where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.

Let's find the angular velocity/

            w= \sqrt{ \frac{200}{0.300} }

            w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

             v= \frac{dx}{dt}

             v= - A w sin (wt +Ф)

             

             0 = -A w sin Ф

            Ф= 0

They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

               4.00 = A cos 25.8t

Speed.

              50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

              Cos² 25.8t = \frac{16}{A^2}

              sin² 25.8t = \frac{3.756}{A^2 }

              1 = \frac{1}{A^2} \ (16 + 3.756)

               A = \sqrt{19.756}

               A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here:  brainly.com/question/14311816

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Match these items. 1. Ca 6 proton 2. H 2O 3 fission 3. nuclear decay 7 element 4. nuclear synthesis 8 electron 5. η atomic numbe
Gnesinka [82]

1. Ca → Element

2. Proton → positive

3. H2O → compound

4. Fission → nuclear decay

5. Fusion → Nuclear synthesis

6. η → Neutron

7. e → electron

8. Atomic number → no of protons in nucleus.

Explanation

1. Ca (Calcium):

Calcium is an element with the atomic number of 20. It is an alkaline earth metal. The 99% of calcium is found in our bodies, in bones, teeth.

2. Proton:

Proton is a subatomic particle and it holds the positive charge. Proton is present in the nucleus of the atom.

3. H2O (water):

Water is a chemical compound and it's chemical formula is H2O. It's called compound as it contains 2 hydrogen and 1 oxygen atoms bonded together through the covalent bond.

4. Fission:

Fission is a process in which large massive unstable nucleus splits into the smaller, less heavier and stable nuclei. The energy is re;eased in the form of radiations during this process. It's called as the radioactive decay.

5. Fusion:

Fusion is opposite of the fission reaction. As in this case the two nuclei combines to form a single large nucleus. That's why it is a nuclear synthesis process.

6. η neutron:

Neutron is a subatomic particle and it is a neutral particle which is located inside the nucleus. n is a symbol used for the neutron.

7. e Electron:

The symbol for electron is e. It's a subatomic particle with negative charge. It is found in the orbits around the nucleus.

8.  Atomic Number:

Atomic number is defined as the number of protons in the nucleus of an atom. IT is represented by Z.

6 0
2 years ago
The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit
sveticcg [70]

Answer:

a)   I = 3.63 W / m² , b)   I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

         I = P / A = ½ ρ v (2π f s_{max})²

         I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

        cte = (½ ρ v 4π² s_{max}²)

         I₀ = cte s² f₀²

        I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

         I = constant (2.20 10³) 2

         I = cte 4.84 10⁶

let's find the relationship of the two quantities

        I / Io = 4.84

        I = 4.84 Io

        I = 4.84 0.750

        I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

        I = cte (f s)²

        I = constant (0.250 10³ 4)²

 

        I = cte 1 10⁶

         

the relationship

        I / Io = 1

        I = Io

        I = 0.750 W / m²

6 0
2 years ago
Hii! help asap. i’ll give brainliest thanks!
o-na [289]

I believe it’s the mass of the box but I don’t no if I’m right

Hope this helped
5 0
3 years ago
A bicyclist is travelling at 25 m/s when he begins to decelerate at -4m/s2 . How fast is travelling after 5 seconds
Umnica [9.8K]

Initial velocity (Vi) = 25 m/s

acceleration (a) = -4 m/s^{2}

time interval (t) = 5 sec

let us assume that final velocity after 5 sec be Vf

As acceleration is constant, we can apply the the equation of motion with constant acceleration i.e. V_{f} = V_{i} + at

Hence, V_{f} = 25 +(-4)(5) = 25 -20 = 5 m/s

so, the velocity of bicyclist will be 5 m/s after 5 sec

7 0
3 years ago
a skydiver jumps out of a plane. describe how gravitational potential energy changes as the skydiver falls. describe how the sky
Oliga [24]

gravitational potential is directly proportional to the height of the object relative to a reference line and is given as

PE = mgh                        

where m = mass of object , g = acceleration due to                    

gravity   and  h = height of the object above the reference line .


as the skydiver falls , its height above the ground decrease and hence the gravitational potential energy of the skydiver decrease.

as per conservation of energy , total energy of the skydiver must remain constant all the time . hence the decrease in potential energy appears as increase in kinetic energy by same amount to keep the total energy constant

KE + PE = Total energy

so as the skydiver falls , it gains speed and hence the kinetic energy of skydiver increase since kinetic energy is directly proportional to the square of the speed.


when the parachute opens, the skydiver experience force in upward which tries to balance the weight of the skydiver. hence the speed of the skydiver decrease until upward force becomes equal to the downward force. hence the kinetic energy decrease just after the parachute opens

4 0
3 years ago
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