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Zarrin [17]
3 years ago
10

A sound wave has a frequency of 425 Hz. What is the period of this wave?

Physics
1 answer:
Serhud [2]3 years ago
8 0
Period is 1/frequency
1/425 = 2.353ms
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A train traveling at 6.4 m/s accelerates at 0.10 m/s 2 over a distance of 100 m. How large is the train’s final velocity?
klasskru [66]

The final velocity of the train at the end of the given distance is 7.81 m/s.

The given parameters;

  • initial velocity of the train, u = 6.4 m/s
  • acceleration of the train, a = 0.1 m/s²
  • distance traveled, s = 100 m

The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;

v² = u² + 2as

v² = (6.4)² + (2 x 0.1 x 100)

v² = 60.96

v = √60.96

v = 7.81 m/s

Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.

Learn more here:brainly.com/question/21180604

4 0
3 years ago
In which region of the sun does energy move as waves that transfer from atom to atom
Artist 52 [7]

Answer:

Radiative zone

4 0
3 years ago
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Please help (choose a b or c)
Dafna11 [192]

I GEUSS A and COM is correct answer to this question

3 0
3 years ago
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A motorbike accelerates from 15m/s to 25m/s in 15 seconds.
RSB [31]

distance traveled by a uniformly accelerated bike is given as

d = \frac{v_f + v_i}{2} (t)

here we know that

v_f = 25 m/s

v_i = 15 m/s

t = 15 s

now we will have from above equation

d = \frac{15 + 25}{2} (15)

d = 20 (15) = 300 m

so it will cover the total distance of 300 m

5 0
3 years ago
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

7 0
3 years ago
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