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jeka94
2 years ago
6

a water wave has a speed of 23.0 meters/second. if the waves frequency is 0.0680 hertz, what is the wavelength

Physics
1 answer:
USPshnik [31]2 years ago
7 0

Answer:

 Using the Fundamental Equation of Wave, we have:

Explanation:

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A train slows from 60m/s to 20m/s in 50s
Dovator [93]

Answer:

a = -4/5 m/s^2

Explanation:

Acceleration = change in velocity / time

change in velocity = final velocity - initial velocity

a = (20 m/s - 60 m/s) / 50 s

a = -40 m/s / 50 s

a = -4/5 m/s^2

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1 year ago
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3 years ago
) Un círculo de 120 cm de radio gira a 600 rpm. Calcula: a) su velocidad angular
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Responder:

20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz

Explicación:

Dado lo siguiente:

Radio (r) del círculo = 120 cm

600 revoluciones por minuto en radianes por segundo

(600 / min) * (2π rad / 1 rev) * (1min / 60seg)

(1200πrad / 60sec) = 20π rad ^ -1

Velocidad angular (w) = 20πrads ^ -1

Velocidad lineal = radio (r) * velocidad angular (w)

Velocidad lineal = (120/100) * 20πrad

Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1

C.) Período (T):

T = 2π / w = 2π / 20π = 0.1 seg

D.) Frecuencia (f):

f = 1 / T = 1 / 0.1

1 / 0,1 = 10 Hz

5 0
3 years ago
Why does the speed of sound depend on air temperature?
IRISSAK [1]
<span>The speed of sound is dependent on how close together the molecules of the transmitting medium is.</span>
4 0
3 years ago
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If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J

So, the required work done is 16 J.

7 0
3 years ago
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