The climate of a town on the coast will have less dramatic changes than a town further inland. Is this statement true or false?

A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of

Wewaii [24]

(1) The wavelength of the wave is 1.164 m.

(2) The velocity of the wave is 23.7 m/s.

(3) The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

<h3> Wavelength of the wave</h3>

A general wave equation is given as;

y(x, t) = A sin(Kx - ωt)

<h3>Velocity of the wave</h3>

v = ω/K

From the given wave equation, we have,

y(x, t) = 0.048 sin(5.4x - 128t)

v = ω/K

where;

ω corresponds to 128
k corresponds to 5.4

v = 128/5.4

v = 23.7 m/s

<h3>Wavelength of the wave</h3>

λ = 2π/K

λ = (2π)/(5.4)

λ = 1.164 m

<h3>Maximum speed of the wave</h3>

v(max) = Aω

where;

A is amplitude of the wave
ω is angular speed of the wave

v(max) = (0.048)(128)

v(max) = 6.14 m/s

Thus, the wavelength of the wave is 1.164 m.

The velocity of the wave is 23.7 m/s.

The maximum speed in the y-direction of any piece of the string is 6.14 m/s.

Learn more about wavelength here: brainly.com/question/10728818

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3 0

2 years ago

If it takes you 10 seconds to move a chair 5 meters across the floor, using a force of 2 Newtons, how much power did you put out

Maru [420]

Answer:

power=work done÷time taken

2×5=10

10÷10=1

ans 1J per second

5 0

2 years ago

1. Two forces act on a box as follows: F= 100 N at 02 = 170. and F2= 75 N at

Setler [38]

Answer:

yfv.

Explanation:

7 0

3 years ago

You are watching an object that is moving in shm. when the object is displaced 0.600 m to the right of its equilibrium position,

larisa [96]

It should take at least 4.50

8 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago