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2 years ago

When a mass of

Physics

1 answer:

Anuta_ua [19.1K]2 years ago

3 0

Answer:

The amplitude of oscillation is <u>0.07698m</u>

Explanation:

The given motion is an example of Simple Harmonic Motion (SHM).

For a simple harmonic spring-block motion the angular frequency (w) is given by -

w = $\sqrt{\dfrac{k}{m} }$ -- 1

,where k=spring constant of the spring and m=mass of the block.

We know that time period of SHM is given by -

T = $\dfrac{2\pi }{w}$ = 0.467

∴ w = $\dfrac{2\pi }{T}$ = $\dfrac{2\pi }{0.467}$ = 13.45 $s^{-1}$

The general equation of motion -

x = A sin(wt+∅) , where A : Amplitude of oscillation

t : time

∅ : phase difference of oscillation

x : Displacement of block

∴ v = Aw cos(wt+∅) (On differentiating) , v : velocity of block

Now ,

Total energy (E) = Kinetic Energy (KE) + Potential Energy (PE)

PE = $\frac{1}{2}$ k $x^{2}$ [Potential energy of spring]

KE = $\frac{1}{2}$ m $v^{2}$

Now,

k=m $w^{2}$ [from 1]

Substituting the value of k,x and v in the equation of KE and PE

We get,

PE = $\frac{1}{2}$ m $w^{2}$ $A^{2} sin^{2}(wt+∅)$

KE = $\frac{1}{2}$ m $A^{2}w^{2}cos^{2}(wt+∅)$

∴ TE = $\frac{1}{2}$ m $w^{2}$ $A^{2} sin^{2}(wt+∅)$ + $\frac{1}{2}$ m $A^{2}w^{2}cos^{2}(w+∅)$

= $\frac{1}{2}$ m $w^{2}A^{2}$

Given , TE = 0.163

∴ 0.163 = $\frac{1}{2}$ m $w^{2}A^{2}$

Substituting m=304g=0.304kg

w = 13.45 $s^{-1}$

We get,

A = 0.07698 m

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A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

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Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

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2 years ago

I wonder if people that say that there ugly is to get attention (not to be mean...)

alexandr1967 [171]

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1 year ago

Read 2 more answers

Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt

Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400)
^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules
$

Hence the required molecules is $13.3\times10^{43} molecules
$

(c)

Write the ideal gas equation as follows:

$PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$

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2 years ago

what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

Approximately $5.11 \times 10^{-19}\; {\rm J}$ .

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

Look up Planck's constant: $h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}$ .

Apply the Rydberg formula to find the wavelength $\lambda$ (in vacuum) of the photon in question:

$\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}$ .

The frequency of that photon would be:

$\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}$ .

Combine this expression with the Rydberg formula to find the frequency of this photon:

$\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}$ .

Apply the Einstein-Planck equation to find the energy of this photon:

$\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}$ .

(Rounded to three significant figures.)

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1 year ago

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

ValentinkaMS [17]

The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

<h3>Acceleration of the box</h3>

The acceleration of the box is calculated as follows;

vf² = vi² + 2as

a = (vf² - vi²)/2s

a = (11.5² - 13²) / (2 x 8.5)

a = -2.16 m/s²

<h3>Time of motion of the box</h3>

The time taken for the box to travel is calculated as follows;

a = (vf - vi)/t

t = (vf - vi) / a

t = (11.5 - 13) / (-2.16)

t = 0.69 s

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P = Fv

P = (ma)(vf - vi)

P = (1 x -2.16) x (11.5 - 13)

P = 3.24 W

Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

Learn more about average power here: brainly.com/question/19415290

#SPJ1

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1 year ago