What is the difference between how transverse waves and longitudinal waves move particles

1 answer:

7 0

A wave on a string is the classic example of a transverse wave. Each part of the string moves up and down while the wave moves from side to side. Transverse waves can not happen in gases because the perpendicular motion is not created by any force.

A Slinky is a great way to visualize longitudinal waves. Each part of the Slinky moves from side to side, just like the wave itself.

Sound waves are longitudinal pressure waves in the air. Water waves involve a combination of transverse and longitudinal waves. The water moves up and down, but also back and forth. Each particle in the water ends up moving in a circular fashion. Earthquakes also have different kinds of waves. The primary waves, called P waves, move with the highest velocity and are transverse waves. Secondary waves, called S waves, are longitudinal waves and occur seconds after the primary waves.

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(b) suppose two telephone poles are 40 ft apart and the length of the wire between the poles is 41 ft. if the lowest point of th

lakkis [162]

Refer to the diagram shown below.

The suspended wire is in the shape of a parabola defined by the equation
y = ax²
where a = a positive constant.
The derivative of y with respect to x is y' = 2ax.

The vertex is at (0,0) and the line of symmetry is x = 0.
The suspended length is 41 ft, therefore half the suspended length is 20.5 ft.
The length between x = 0 and x = 20 is given by
$\int _{0}^{20} \sqrt{1+[y'(x)]^{2}} \, dx = \int_{0}^{20} \sqrt{1+4a^{2}x^{2}} \, dx =20.5$

Because we do not know the value of a, we shall find it numerically.
Define the function
$f(a) = \int _{0}^{20} \sqrt{1+4a^{2}x^{2}} \, dx - 20.5 = 0$
The plot for f(a) versus a yields an approximate solution (from Matlab) of a = 0.01 (shown in the figure).

Therefore
y = 0.01x²
When x = 20 ft, h = 0.01(400) = 4 ft
Because the vertex of the parabola is 19 ft above ground, the support points for the wire are 19 + h = 23 ft above ground.

Answer: 23.00 ft

7 0

3 years ago

A book prone to air resistance is released from rest 300 m

yaroslaw [1]

Answer:

Approximately $73\%$ .

(Assuming that $g = \rm 9.81\; m \cdot s^{-2}$ .)

Explanation:

The mechanical energy of an object is the sum of its potential energy and its kinetic energy. It will be shown that the exact mass of this object doesn't matter. For ease of calculation, let $m(\text{book})$ represent the mass of the book.

The initial potential energy of the book is

$\begin{aligned}U(300\; \text{m}) &= m(\text{book}) \cdot g \cdot \Delta h + U(0\; \text{m}) \cr &=(9.81 \times 300) \cdot m(\text{book})\cr &= \left(2.943\times 10^3\right) \cdot m(\text{book})\end{aligned}$ .

The book was initially at rest when it was released. Hence, its initial kinetic energy would be zero. Hence, the initial mechanical energy of the book-Earth system would be $(2.943\times 10^3) \cdot m(\text{book})$ .

When the book was about to hit the ground, its speed is $\rm 40\; m \cdot s^{-1}$ . Its kinetic energy would be:

$\begin{aligned} \text{KE} &= \frac{1}{2} \, m(\text{book}) \cdot v^{2} \cr &= \left(\frac{1}{2} \times 40^2\right)\cdot m(\text{book}) \cr &= \left(8.00\times 10^2\right)\cdot m(\text{book})\end{aligned}$ .

The question implies that the potential energy of the book near the ground is zero. Hence, the mechanical energy of the system would be $\left(8.00\times 10^2\right)\cdot m(\text{book})$ when the book was about to hit the ground.

The amount of mechanical energy lost in this process would be equal to:

$\begin{aligned}&\left(2.943\times 10^3\right) \cdot m(\text{book}) - \left(8.00\times 10^2\right)\cdot m(\text{book}) \cr &=\left(2.143\times 10^3\right)\cdot m(\text{book})\end{aligned}$ .

Divide that with the initial mechanical energy of the system to find the percentage change. Note how the mass of the book, $m(\text{book})$ , was eliminated in this process.

$\begin{aligned}&\frac{\left(2.143\times 10^3\right)\cdot m(\text{book})}{\left(2.943\times 10^3\right) \cdot m(\text{book})}\times 100\% \cr &= \frac{2.143\times 10^3}{2.943\times 10^3}\times 100\% \cr & \approx 73\%\end{aligned}$ .