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2 years ago

The period (T) of an oscillating wave is 1/5s. What happens to the frequency (f) of the wave if T increases to 1/2s

1 answer:

6 0

Frequency = 1/T
as the 5 is reduced, frequency is increase.
as 1 whole wave travels through a point in a lesser time now

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The position of an object in simple harmonic motion is defined by the function y = (0.50 m) sin (πt/2). Determine the maximum sp

Gnoma [55]

The maximum speed of the object under simple harmonic motion is 0.786 m/s.

The given parameters:

Position of the particle, y = 0.5m sin(πt/2)

<h3>Wave equation for simple harmonic motion;</h3>

y = A sin(ωt + Ф)

where;

A is the amplitude = 0.5 m
ω is the angular speed = π/2

The maximum speed of the object is calculated as follows;

$V_{max} = A \omega\\\\V_{max} = 0.5 \times \frac{\pi}{2} = \frac{\pi}{4} \ m/s = 0.786 \ m/s$

Thus, the maximum speed of the object under simple harmonic motion is 0.786 m/s.

Learn more about simple harmonic motion here: brainly.com/question/17315536

5 0

2 years ago

What is the velocity if your going 105 miles every 10 minutes

Afina-wow [57]

The speed is 10.5 miles per minute, or 630 miles per hour.
We don't know the velocity, because you didn't tell us anything about the direction I'm going.

8 0

2 years ago

Read 2 more answers

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed

kogti [31]

Answer:

a) $y_{first}=5.3mm$

b) $y_{second}=10.6-5.3 =5.3 mm$

Explanation:

The width of the central bright in this diffraction pattern is given by:

$y=\frac{m\lambda D}{a}$ when m is a natural number.

here:

m is 1 (to find the central bright fringe)
D is the distance from the slit to the screen
a is the slit wide
λ is the wavelength

So we have:

$y_{first}=\frac{633*10^{9}*3.35}{0.0004}$

$y_{first}=5.3mm$

Now, if we do m=2 we can find the distance to the second minima.

$y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}$

$y_{2}=10.6 mm$

Now we need to subtract these distance, to get the width of the first bright fringe :

$y_{second}=10.6-5.3 =5.3 mm$

I hope it heps you!

4 0

2 years ago

A mass m is tied to an ideal spring with force constant k and rests on a frictionless surface. The mass moves along the x axis.

7nadin3 [17]

Answer: $x=\frac{x_m}{\sqrt{2}}$

Explanation:

Given

initially mass is stretched to $x_m$

Let k be the spring Constant of spring

Therefore Total Mechanical Energy is $\frac{kx_m^2}{2}$

Position at which kinetic Energy is equal to Elastic Potential Energy

$K=\frac{mv^2}{2}$

$U=\frac{kx^2}{2}$

it is given

$k=U$

thus $2U=\frac{kx_m^2}{2}$

$2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}$

$2x^2=x_m^2$

$x=\frac{x_m}{\sqrt{2}}$

3 0

3 years ago

The path that thermal energy follows is from ___________ to ___________

Helga [31]

Warmer to colder objects

5 0

3 years ago