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3 years ago

What happens to a wave when reflected off a node

1 answer:

7 0

A light wave travelling in air that is reflected by a glass barrier will undergo a 180° phase change, while light travelling in glass will not undergo a phase change if it is reflected by a boundary with air. ... The phase changes that take place upon reflection play an important part in thin film interference.

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How would you explain social control to someone who has no idea about our society?

Natasha2012 [34]

You can try to explain it by using a parallel between their and your societies .

7 0

3 years ago

An object is given a very small amount of charge. Which of the following could

spayn [35]

5.4*10^-19 C

Explanation:

For the purposes of this question, charges essentially come in packages that are the size of an electron (or proton since they have the same magnitude of charge). The charge on an electron is -1.6*10^-19

Therefore, any object should have a charge that is a multiple of the charge of an electron - It would not make sense to have a charge equivalent to 1.5 electrons since you can't exactly split the electron in half. So the charge of any integer number of electrons can be transferred to another object.

Charge = q(electron)*n(#electrons)

Since 5.4/1.6 = 3.375, we know that it can not be the right answer because the answer is not an integer.

If you divide every other option listed by the charge of an electron, you will get an integer number.

(16*10^-19 C)/(1.6*10^-19C) = 10

(-6.4*10^-19 C)/(1.6*10^-19C) = -4

(4.8*10^-19 C)/(1.6*10^-19C) = 3

(5.4*10^-19 C)/(1.6*10^-19C) = 3.375

(3.2*10^-19C)/(1.6*10^-19C) = 2

etc.

I hope this helps!

3 0

3 years ago

The vertical force f acts downward at A on the two membered frames. Determine the magnitude of the two components of F directed

galben [10]

Answer:

The magnitude will be "353.5 N". A further solution is given below.

Explanation:

The given values is:

F = 500 N

According to the question,

In ΔABC,

⇒ $\angle BCA = (90-30)$

⇒ $=60^{\circ}$

then,

⇒ $\angle BAC=(180-45-60)$

⇒ $=75^{\circ}$

Now,

The corresponding angle will be:

⇒ $\angle FAC=60^{\circ}$

⇒ $\angle FAB=70+60$

⇒ $=135^{\circ}$

Aspect of F across the AC arm will be:

= $F\times cos(60)$

On putting the values of F, we get

= $500\times (.5)$

= $200 \ Newton$

Component F along the AC (in magnitude) will be:

= $F\times cos(135)$

= $500\times (-.707)$

= $-353.5 \ N \$

4 0

3 years ago

A pendulum on plant X is 0.25 meters long and the period is 4 seconds. What is “g” on Planet X?

Monica [59]

Answer:

The value of g = 0.6168 m/s².

Explanation:

Given that,

On a planet X,

Length of the pendulum(L) = 0.25 meters,

Time period of the pendulum(T) = 4 seconds.

We have to find the 'g' value on the planet.

The 'g' value on a planet can be found by a pendulum with help of the formula,

T = 2π × $\frac{\sqrt{L} }{\sqrt{g} }$

From this, g = 4π² × $\frac{L}{T^{2} }$

Using the above formula and substituting the values,we get,

g = 0.6168 m/s².

7 0

3 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago