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FinnZ [79.3K]
1 year ago
9

A sinusoidal wave in a string is described by the wave functiony=0.150 sin (0.800x - 50.0t)where x and y are in meters and t is

in seconds. The mass per length of the string is 12.0 g/m(a) Find the maximum transverse acceleration of an element of this string
Physics
1 answer:
AleksAgata [21]1 year ago
6 0

The maximum  transverse acceleration of an element of this string is 375 m/s

<h3>What is the  maximum transverse acceleration?</h3>

For a transverse wave, the velocity is known to be perpendicular to the direction of the propagation of the particular wave. To get the acceleration, what is necessary is to take the partial  derivative as it compares to time and the velocity.

With this in mind, we would have

transverse acceleration

a(t) = d²y/dt² = -0.150*50.02sin(0.8x - 50t)

= 0.150*502 = 375 m/s2

Hence we can say that the maximum acceleration of the element on the string is given as 375 m/s²

Read more on acceleration here

brainly.com/question/605631

#SPJ4

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A car has a mass of 975 kg. by pushing on the car, evan increses its speed from 0.5 m/s to 3 m/s. what impulse did evan apply to
jarptica [38.1K]
Impulse = change in momentum = F x dt
Where F is sufficiently large and dt is very small(tending to zero).

Therefore impulse = mass x final velocity - mass x initial velocity
                                = 975 x 3 - 975 x 0.5
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Hence option C is correct.

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3 years ago
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Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.3 m from the slits. The center of the 3
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Answer:

a) f = 4.76 10¹⁴ Hz, b) d = 2.73 10⁻⁴ m, c) θ = 6.923 10⁻³ rad

Explanation:

a) In this problem the frequency of light is asked, let's use the relationship between the speed of the wave, its wavelength and its frequency

           c = λ f

           f = c /λ

           f = \frac{3 \ 10^8}{630 \ 10^{-9}}

           f = 4.76 10¹⁴ Hz

b) slit separation (d)

the expression for the constructive interference of the double-slit experiment is

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let's use trigonometry

          tan θ = y / L

          tan θ = \frac{sin \theta}{cos \theta}

in general the angles are small, so we can approximate

          tan θ = sin θ

          tan θ = y/L

we substitute

          d y / L = m λ

          d = m L λ / y

we calculate

          d = 3  1.3  630 10⁻⁹ /0.90 10⁻²

          d = 2.73 10⁻⁴ m

c) the angle

           tan θ = y / L

           θ = tan⁻¹ y / L

           θ = tan⁻¹ 0.9 10⁻² / 1.3

           θ = tan⁻¹ 6,923 10⁻³

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           θ = 6.923 10⁻³ rad

   

4 0
3 years ago
See Conceptual Example 6 to review the concepts involved in this problem. A 12.0-kg monkey is hanging by one arm from a branch a
labwork [276]

Answer:

(a)  24.56 N

(b) 142.28 N

Explanation:

(a)

The designation assigned to something like the net force pointed toward the middle including its circular route seems to be the centripetal force. The net stress only at lowest point constitutes of the strain throughout the arm projecting upward towards the middle as well as the weight pointed downwards either backwards from the center.

The centripetal function is generated from either scenario by Equation:

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⇒  T = mg + Fc

⇒      =12\times 9.81+24.56

⇒      =142.28 \ N

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