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weeeeeb [17]
2 years ago
7

Consider a standing wave on a string. What is the distance between two adjacent nodes in terms of the wavelength λ of the standi

ng wave?.
Physics
1 answer:
amid [387]2 years ago
3 0

The distance between the two adjacent nodes = λ/2.

<h3>What is Wavelength?</h3>

A periodic wave's wavelength is its spatial period, or the length over which its form repeats. It is a property of both travelling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings. The spatial frequency is the reciprocal of wavelength. The Greek letter lambda (λ) is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

The distance between the two adjacent nodes = λ/2.

for the standing wave ,the distance between any two adjacent nodes or antinodes is 1/2 λ.

to learn more about the wavelength go to - brainly.com/question/6297363

#SPJ4

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What will be the average velocity of a body falling in free fall on Earth for 3 s?
SpyIntel [72]

Answer:

29.4m/s

Explanation:

Given parameters:

Time  = 3s

Unknown:

Average velocity  = ?

Solution:

To solve this problem, we use the expression below:

      v  = u + gt

v is the average velocity

u is the initial velocity  = 0m/s

g is the acceleration due to gravity  = 9.8m/s²

t is the time

So;

        v  = 0 + (9.8 x 3)  = 29.4m/s

6 0
3 years ago
A 40 kg boy is moving upwards upwards in a lift with an acceleration of 2 m /s ^2 what would be the weight felt by him if measuu
neonofarm [45]

Answer:

<h2> 48kg</h2>

f = ma \\

w \:  =  \frac{f}{g}  \\

Explanation:

f \:  = ma \\ f - mg = ma \\ f = ma + mg \\ f = 40 \times 2 + 40 \times 10 \\ f = 480

w =  \frac{f}{g}  \\w =   \frac{480}{10}  \\ w = 48kg

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3 years ago
When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

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The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

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