Physics, 14.04.2020 17:26, MajentaSnow11601

A 0.227-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.279 m and x2 = 0.469 m. The period of oscillation is 0.609 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.

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Answers

a) $f = 1.642\,hz$ , b) $\bar x = 0.095\,m$ , c) $A = 0.374\,m$ , d) $v_{max} = 3.859\,\frac{m}{s}$ , e) $a_{max} = 39.809\,\frac{m}{s^{2}}$ , f) $k = 24.162\,\frac{N}{m}$ , g) $E = 1.690\,J$

Explanation:

a) The frequency is the reciprocal of the period of oscillation:

$f = \frac{1}{T}$

$f = \frac{1}{0.609\,s}$

$f = 1.642\,hz$

b) The equilibrium position is the average of the extreme points:

$\bar x = \frac{-0.279\,m +0.469\,m}{2}$

$\bar x = 0.095\,m$

c) The amplitude is the absolute of the substraction of the equilibrium position from any of the extreme points:

$A = |0.469\,m - 0.095\,m|$

$A = 0.374\,m$

d) The angular frequency is:

$\omega = 2\pi \cdot f$

$\omega = 2\pi \cdot (1.642\,hz)$

$\omega \approx 10.317\,\frac{rad}{s}$

The maximum speed is:

$v_{max} = \omega \cdot A$

$v_{max} = (10.317\,\frac{rad}{s} )\cdot (0.374\,m)$

$v_{max} = 3.859\,\frac{m}{s}$

e) The maximum acceleration is:

$a_{max} = \omega^{2}\cdot A$

$a_{max} = (10.317\,\frac{rad}{s} )^{2}\cdot (0.374\,m)$

$a_{max} = 39.809\,\frac{m}{s^{2}}$

f) The force constant is:

$k = \omega^{2}\cdot m$

$k = (10.317\,\frac{rad}{s} )^{2}\cdot (0.227\,kg)$

$k = 24.162\,\frac{N}{m}$

g) The total mechanical energy is:

$E = \frac{1}{2}\cdot k \cdot A^{2}$

$E = \frac{1}{2}\cdot (24.162\,\frac{N}{m} )\cdot (0.374\,m)^{2}$

$E = 1.690\,J$

answered: Guest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

answered: Guest

the grams of oxygen gas that can be produced is 20.6 grams of o₂

calculation

2liclo₃ → 2 licl + 3 o₂

step 1: find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5 +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281 moles

step 2: use the mole ratio to calculate the moles of o2

from equation above, liclo₃: o₂ is 2: 3

therefore the moles of o₂=0.4281 moles x3/2 =0.6422 moles

step 3: find mass of o₂

mass= moles x molar mass

from periodic table the molar mass of o₂ = 16 x2= 32 g/mol

mass=0.6422 moles x 32 g/mol = 20.6 grams of o₂

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A 0.227-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points...