What is the potential energy of a 500 N weight before it falls 10 m ?

You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you

vekshin1

Answer:

Approximately $5.9\; {\rm m\cdot s^{-1}}$ (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

the antinode at the end of the spring that is being shaken, and
the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

node at the end that is being held still,
antinode (as mentioned in the question),
node (inferred, not mentioned in the question), and
antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, $(1/2)$ ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, $(1/4)$ ) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is $\lambda$ , the length of this spring would be:

$\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda$ .

The question states that the length of this coiled spring is $7.2\; {\rm m}$ . In other words, $(3/4) \, \lambda = 7.2\; {\rm m}$ . The wavelength of this wave would be $(7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}$ .

The frequency $f$ of this wave is the number of cycles in unit time:

$\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}$ .

Hence, the speed $v$ of this wave would be:

$\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}$ .

3 0

2 years ago

The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.

Sholpan [36]

The frequency of rotation of Mars is 0.0000113 Hertz.

Given the following data:

Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

Conversion:

1 day = 24 hours

24 hours to minutes = $60$ × $24$ = $1440$ minutes