Compare the 3 different types of waves

Oduvanchick [21]

Bacteria <span>are very small </span>organisms<span>.</span>

6 0

4 years ago

A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 2

Tom [10]

the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:

The range of motion is: A = 4.44 cm

<h3>Oscillatory movement.</h3>

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

x = A cos (wt + Ф)

w² = k/m

where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.

Let's find the angular velocity/

w= $\sqrt{ \frac{200}{0.300} }$

w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

v= $\frac{dx}{dt}$

v= - A w sin (wt +Ф)

0 = -A w sin Ф

Ф= 0

They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

4.00 = A cos 25.8t

Speed.

50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

Cos² 25.8t = $\frac{16}{A^2}$

sin² 25.8t = $\frac{3.756}{A^2 }$

1 = $\frac{1}{A^2} \ (16 + 3.756)$

A = $\sqrt{19.756}$

A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here: brainly.com/question/14311816

4 0

2 years ago

A 16.0 Ω, 13.0 Ω, and 7.00 Ω resistor are connected in parallel to an emf source. A current of 6.00 A is in the 13.0 Ω resistor.

Darya [45]

Answer:3.54ohms

Explanation: connection in parallel

1/Rt= 1/R1+1/R2+1/R3

1/Rt= 1/16+1/13+1/7

1/Rt= 91+112+208/1456

1/Rt= 411/1456

411Rt= 1456

Rt= 1456/411

Rt= 3.54ohms

3 0

4 years ago

Read 2 more answers

An LED operation at 850 nm center wavelength has a spectral width of 45 nm. What is the pulse spreading in ns/km

Papessa [141]

Answer:

$\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}$

Explanation:

From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.

Hence, the material dispersion is $\dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )$

Now, using the pulse spread formula:

$\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda$

$\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ ) \times (45 \ nm)$

Thus, the pulse spreading as a result of material dispersion is: $\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}$

3 0

3 years ago

At what time was the person at a position of 0m?

Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

= (0 x 1) + 1/2 (0 x 1 x 1)

= (0) + 1/2 (0)

= 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

= 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1) = 10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0

3 years ago