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Scilla [17]
2 years ago
7

Transverse waves on a string have wave speed v=8.00 m/s, amplitude A=0.0700m, and direction, and at t=0 the x-0 end of the wavel

ength -0.320m. The waves travel in the -x string has its maximum upward displacement. a) Find the frequency, period and wave number of these waves b) Write a wave function describing the wave c) Find the transverse displacement of a particle at x=0.360m at time t=0.150 d) How much time must elapse from the instant in part (c) until the particle at x-0.360 m next has maximum upward displacement?
Physics
1 answer:
Vilka [71]2 years ago
6 0

Answer:

a. frequency = 25 Hz, period = 0.04 s , wave number = 19.63 rad/m

b. y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

c.  0.0496 m

d. 0.03 s

Explanation:

a. Frequency, f = v/λ where v = wave speed = 8.00 m/s and λ = wavelength = 0.320 m

f = v/λ = 8.00 m/s ÷ 0.320 m = 25 Hz

Period, T = 1/f = 1/25 = 0.04 s

Wave number k = 2π/λ = 2π/0.320 m = 19.63 rad-m⁻¹

b. Using y = Asin(kx - ωt) the equation of a wave

where y = displacement of the wave, A = amplitude of wave = 0.0700 m and ω = angular speed of wave = 2π/T = 2π/0.04 s = 157.08 rad/s

Substituting the variables into y, we have

y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

c. When x = 0.360 m and t = 0.150 s, we substitute these into y to obtain

y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

y = (0.0700 m)sin[(19.63 rad/m × 0.360 m) - (157.08 rad/s × 0.150 s)]

y = (0.0700 m)sin[(7.0668 rad) - (23.562 rad)]

y = (0.0700 m)sin[-16.4952 rad]

y = (0.0700 m) × 0.7084

y = 0.0496 m

d. For the particle at x = 0.360 m to reach its next maximum displacement, y = 0.0700 m at time t. So,

y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]

0.0700 m = (0.0700 m)sin[(19.63 rad/m × 0.360 m) - (157.08 rad/s)t]

0.0700 m = (0.0700 m)sin[(7.0668 rad - (157.08 rad/s)t]  

Dividing through by 0.0700 m, we have

1 = sin[(7.0668 rad - (157.08 rad/s)t]

sin⁻¹(1) = 7.0668 rad - (157.08 rad/s)t  

π/2 = 7.0668 rad - (157.08 rad/s)t

π/2 - 7.0668 rad = - (157.08 rad/s)t

-5.496 rad = - (157.08 rad/s)t

t = -5.496 rad/(-157.08 rad/s) = 0.03 s

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Lena [83]

Answer:

  E =  k Q    1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

              dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

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We integrate

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We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

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We replace the density

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8 0
2 years ago
An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's
ivolga24 [154]
1) Focal length

We can find the focal length of the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i} (1)
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d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}
from which we find
f=- \frac{36}{7} cm=-5.1 cm

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

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3 0
2 years ago
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m  

0.80 m  

1.25 m

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0
2 years ago
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Answer:

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Explanation:

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kx² = mv²

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v² = 75 / 0.05

v² = 1500

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8 0
2 years ago
A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text
antoniya [11.8K]

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8 0
2 years ago
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