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Dmitriy789 [7]
3 years ago
15

You attach a 1.90 kg mass to a horizontal spring that is fixed at one end. You pull the mass until the spring is stretched by 0.

400 m and release it from rest. Assume the mass slides on a horizontal surface with negligible friction. The mass reaches a speed of zero again 0.600 s after release (for the first time after release). What is the maximum speed of the mass (in m/s)
Physics
1 answer:
Gre4nikov [31]3 years ago
6 0

Answer:

2.09 m/s

Explanation:

As the  spring is stretched initially , and the mass released from rest i.e v=0. Also, The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.600 s. This illustrates half oscillation of the system.

Therefore, for the period of a full oscillation of the system

T= 2t => 2(0.6)=> 1.2 s

As the frequency is the reciprocal of the period, we have

f= 1/T => 1/1.2

f= 0.833 Hz

The angular frequency'ω' is given by,

ω= 2πf => 2π x 0.833=>5.23 rad/s

For the maximum velocity of the object  in a spring-mass system:

V(_{max} )= Aω

where A is the amplitude of the oscillation. As here, the amplitude of the motion corresponds to the initial displacement of the object (A=0.400 m)

V(_{max} )= 0.4 x 5.23 =>2.09 m/s

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Kinetic energy = 1/2 * mass * velocity^2

In this case,
KE = 1/2 * 1569 kg * (15 (m/s))^2 = 176,5 kN
8 0
3 years ago
A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to
dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence,i=30^{\circ}

Angle of refraction,r=19.24^{\circ}

a.Refractive index of air,n_1=1

We know that

n_2sinr=n_1sini

n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517

b.Wavelength of red light in vacuum,\lambda=632.8nm=632.8\times 10^{-9} m

1nm=10^{-9} m

Wavelength in the solution,\lambda'=\frac{\lambda}{n_2}

\lambda'=\frac{632.8}{1.517}=417nm

c.Frequency does not change .It remains same in vacuum and solution.

Frequency,\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}

Where c=3\times 10^8 m/s

Frequency,\nu=4.74\times 10^{14}Hz

d.Speed in the solution,v=\frac{c}{n_2}

v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s

5 0
3 years ago
Based on the data in the table...especially the provided melting points, which two substances are
ArbitrLikvidat [17]

The two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.

<h3 /><h3 /><h3>What is a covalent Bond?</h3>
  • A covalent bond is a type of chemical bond that involves the sharing of pairs of electron between atoms.

Examples of compounds with covalent bond include the following;

  • Distilled water
  • Sucrose
  • Ethanol

Olive oil is a mixture not a compound

Sodium Chloride & Potassium lodide are examples of ionic bond.

Thus, the two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.

Learn more about covalent bonds here: brainly.com/question/12732708

7 0
2 years ago
In an eslastic ,the momentum is_______and the mechanical energy is​
DIA [1.3K]

Answer:

In an elastic collision, the momentum is conserved and the mechanical energy is conserved too.

Explanation:

There are two types of collisions:

- Elastic collision: in an elastic collision, the total momentum before and after the collision is conserved; also, the total mechanical energy before and after the collision is conserved.

- Inelastic collision: in an inelastic collision, the total momentum before and after the colllision is conserved, while the total mechanical energy is not conserved (in fact, part of the energy is converted into other forms of energy such that thermal energy, due to the presence of frictional forces)

3 0
3 years ago
Answer the question based on this waveform.
Nuetrik [128]

Answer:

Cannot be determined from the given information

Explanation:

Given the following data;

Velocity = 24 m/s

Period = 3 seconds

To find the amplitude of the wave;

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

t is the time period measured in seconds.

ϕ is the phase angle.

Hence, the information provided in this exercise isn't sufficient to find the amplitude of the waveform.

However, the given parameters can be used to calculate the frequency and wavelength of the wave.

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