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Gala2k [10]
3 years ago
7

An air-track glider attached to a spring oscillates between the 13.0 cm mark and the 63.0 cm mark on the track. The glider compl

etes 15.0 oscillations in 32.0 s . What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?
Physics
1 answer:
LuckyWell [14K]3 years ago
6 0

Answer:

(A) = 2.13 s

(B) = 0.47 Hz

(C) = 0.25 m

(D) = 0.74 m/s

Explanation:

number of oscillations (n) = 15

time (t) = 32 secs

start point (L1) = 13 cm = 0.13 m

End point (L2) = 63 cm = 0.63 m

(A) period = time / number of oscillation

           = 32 / 15 = 2.13 s

(B) frequency = 1 / period

              = 1 / 2.13 = 0.47 Hz

(C) Amplitude = 0.5 ( L2 - L1 )

                           = 0.5 ( 0.63 - 0.13 )

                          = 0.25 m

(D) max speed = (2π / T) x A

                        =  (2π / 2.13) x 0.25

                       = 0.74 m/s

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Arsenic diffusion in Si: Arsenic is diffused into a thick slice of silicon with no previous arsenic in it at 1100ºC. If the surf
Rainbow [258]

Answer:

Diffusion time is 7.42 h

Solution:

As per the question:

Temperature, T = 1100^{\circ}C

Surface concentration of arsenic, C_{S} = 5.0\times 10^{18}\ atoms/cm^{3}

Surface concentration below Silicon surface, C_{x} = 1.5\times 10^{16}\ atoms/cm^{3}

D = 3.0\times 10^{- 14}\ cm^{2}/s

x = 1.2\mu m = 1.2\times 10^{- 4}\ cm

Initial concentration at t = 0, C_{o} = 0

Now, by using Flick's second eqn:

\frac{C_{S} - C_{x}}{C_{x} - C_{o}} = erf(\frac{x}{\sqrt{Dt}})

Thus by putting appropriate values:

\frac{5.0\times 10^{18} - 1.5\times 10^{16}}{5.0\times 10^{18}} = erf(\frac{1.2\times 10^{- 4}}{2\sqrt{3.0\times 10^{- 14}t}})

0.997 = erf(\frac{364.4}{\sqrt{t}})              (1)

Now,

erf(z) = 0.997

Now, from error function values tabulation:

For z = 2.0, erf(z) = 0.998

For z = 2.2, erf(z) = 0.995

Now,

With the help of linear interpolation method:

\frac{z - 2}{2.2 - 2.0} = \frac{0.997 - 0.995}{0.998 - 0.995}

z = 2.12

Now, using eqn (1) and above value:

\frac{364.4}{\sqrt{t}} = 2.12

t = (\frac{364.4}{2.12})^{2} = 26700 s

t = \frac{26700}{3600} = 7.42\ h

7 0
3 years ago
States of Matter:
GREYUIT [131]
Solid because the particles vibrate in place
5 0
3 years ago
A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi
stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

v=\sqrt{\frac{TL}{M}} (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

v= \lambda f (2)

We can equate expression (1) and (2):

\sqrt{\frac{TL}{M}}=\lambda f

Solving for T

T= \frac{M(\lambda f)^2}{L} (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic N_{harmonic} is:

\lambda_{harmonic}=\frac{2l}{N_{harmonic}}

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m

We can now find T on (3) using all the values we have:

T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}

T=75.22 N

3 0
4 years ago
Que carro ese que está en la foto
Taya2010 [7]
Travis Scott!3&;8284$28&:!;&29395
7 0
3 years ago
Read 2 more answers
The current supplied by a battery in a portable device is typically about 0.151 A. Find the number of electrons passing through
butalik [34]

Answer:

n = 1.7*10²² electrons.

Explanation:

  • As the current, by definition, is the rate of change of charge, assuming that the current was flowing at a steady rate of .151 A during the 5 hours, we can find the total charge that passed perpendicular to the cross-section of the circuit, as follows:

       I =\frac{\Delta q}{\Delta t} \\ \\ \Delta q = I* \Delta t \\ \\ \Delta t = 5hs*\frac{3600s}{1h} = 18000 s

       ⇒ Δq = I * Δt = 0.151 A * 18000 s = 2718 C

  • As this charge is carried by electrons, we can express this value as the product of the elementary charge e (charge of  a single electron) times the number of electrons  flowing during that time, as follows:

         Δq = n*e

  • Solving for e:

        n = \frac{\Delta q}{e} =\frac{2718C}{1.6-19C} = 1.7e22    electrons.

6 0
3 years ago
Read 2 more answers
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