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sleet_krkn [62]

3 years ago

14

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di

splaced from its equilibrium position and undergoes simple harmonic oscillations when released. What is the period of the oscillations?

1 answer:

AleksandrR [38]3 years ago

4 0

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

$F=kx$

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

$k= \frac{F}{x}$

$k= \frac{mg}{x}$

$k= \frac{(0.35)(9.8)}{12*10^{-2}}$

$k = 28.58N/m$

Perioricity in an elastic body is defined by

$T = 2\pi \sqrt{\frac{m}{k}}$

Where,

m = Mass

k = Spring constant

$T = 2\pi \sqrt{\frac{0.35}{28.58}}$

$T = 0.685s$

Therefore the period of the oscillations is 0.685s

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Explain how the information from the stimuli is communicated to the brain.

irakobra [83]

Answer:

Answer:Neurons communicate through an electrochemical process. Sensory receptors interact with stimuli such as light, sound, temperature, and pain which is transformed into a code that is carried to the brain by a chain of neurons. Then systems of neurons in the brain interpret this information.

Answer:Neurons communicate through an electrochemical process. Sensory receptors interact with stimuli such as light, sound, temperature, and pain which is transformed into a code that is carried to the brain by a chain of neurons. Then systems of neurons in the brain interpret this information.Explanation:

bro edit it yrself

6 0

3 years ago

Read 2 more answers

A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?

serious [3.7K]

Answer : The wavelength of photon is, $4.63\times 10^{2}nm$

Explanation : Given,

Energy of photon = $4.29\times 10^{-19}J$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength of photon = ?

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}$

$\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm$

Conversion used : $1nm=10^{-9}m$

Therefore, the wavelength of photon is, $4.63\times 10^{2}nm$

6 0

3 years ago

A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.

SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of $x_0 = 5.0 m$

- at t = 0, the initial velocity of the motorcyclist is $v_0 = 15.0 m$ east

- The acceleration of the motorcyclist is constant and it is $a=4.0 m/s^2$

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

$x(t)=x_0 + v_0t + \frac{1}{2}at^2$

where t is the time.

Substituting t = 2.0 s, we find the position:

$x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m$

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

$v(t)=x'(t)=v_0 + at$

where:

$v_0=15.0 m/s$ is the initial velocity

$a=4.0 m/s^2$ is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

$t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s$

And therefore, the position at t = 2.5 s is:

$x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0

3 years ago

leva [86]

Answer:

Option C

Maximum potential energy is at point R.

Explanation:

Potential energy is a product of mass, acceleration due to gravity and height ie

PE=mgh where PE is the potential energy, m is mass of an object, g is acceleration due to gravity whose value is normally taken as 9.81 and h is height. Since at point R we have the maximum height, the potential energy will be highest at this point.

3 0

3 years ago

The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,

madam [21]

Answer:

<u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

E = hf

E = h.v/λ

λ = h/(mv)

E = (1/2)mv²

Where:

E = kinetic energy of the particle
λ = de-Broglie wavelength
m = mass of the particle
v = speed of the particle
h = Planck constant

<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

$\dfrac{m_\alpha}{m_p}=4$

For the kinetic energies you find:

$\dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}$

$\dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4$

Thus:

$\dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}$

$m_\alpha v_\alpha=m_pv_p$

From de-Broglie equation, λ = h/(mv)

$\dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1$

5 0

3 years ago