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3 years ago

What is an Amplitude

2 answers:

5 0

DescriptionThe amplitude of a periodic variable is a measure of its change in a single period. There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values. In older texts the phase is sometimes called the amplitude.

dangina [55]3 years ago

3 0

Amplitude is a measurement of the magnitude of displacement (or maximum disturbance) of a medium from its resting state, as diagramed in the peak deviation example below (it can also be a measurement of an electrical signal's increased or decreased strength above or below a nominal state).

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Which area of earth is most similar to the suns convention zone

julsineya [31]

The area of the Earth that is most similar to the Sun's convection zone would be the mantle. The convection zone of the sun is its outermost layer where heat transfer by convection happens which is similar to the Earth's mantle. It would be the crust because it is the outer most layer.

3 0

3 years ago

Mr. Smith used 606 kWh for the month. If the service charge was $61.37 (see back page of the bill), what is the approximate char

Murrr4er [49]

Answer:

.10/KWh

Explanation:

divide 606 by 61.37 and you get .1012...

7 0

3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

In deep space there is very little friction once they are launched into a probe into deep space where there are no external forc

BabaBlast [244]

Continue on the momentum it has. The probe will continue in the same direction it is moving because there are no forces to act against it. I think this is the answer you are looking for...?

3 0

4 years ago

Read 2 more answers

How are slow twitch fibers and fast twitch fibers different?

Vlad1618 [11]

Slow-twitch muscles help enable long-endurance feats such as distance running.
fast-twitch muscles fatigue faster but are used in powerful bursts of movements like sprinting.

4 0

3 years ago