Exercise provides a healthy outlet for feelings, which helps improve

What does it mean to say that a wave is a cyclic, or periodic, disturbance

Ivenika [448]

Wave is a disturbance or energy that propagate through medium from one point to other point

So basically it is a flowing energy that flows into the medium and hence medium particles start oscillating about their mean position to and fro.

This motion of medium particles or this to and fro motion is about their mean position and this will always be cyclic or periodic motion

This means the disturbance or energy continuously flow through the medium such that it will change the position of medium particle and this will be cyclic in order

For an example

$y = A sin(\omega t + kx + \phi)$

so here above equation of wave is a travelling wave in which displacement of medium particle from its mean position is given by "y"

Now we can see that this disturbance depends upon the sine function and it will repeat its same position after every 2 pi time interval as it is cyclic function for this value

Due to this phenomenon of repeatation of its same position we can say that it is disturbance of wave is cyclic.

3 0

2 years ago

Which of the following is not an example of centripetal acceleration?

Amanda [17]

An apple falling to the ground is not an example of centripetal acceleration.

5 0

3 years ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and

Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

$\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m$

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}$ (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

$d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m$

hence, the displacement of the airplane is 3.45*10^4 m

6 0

2 years ago

A 2011 Porsche 911 Turbo S goes from 0-27 m/s in 2.5 seconds. What is the car's acceleration?

Natalka [10]

Answer:

-10.8m/s^2

Explanation:

a=change in velocity/change in time

-27 m/s/2.5=10.8m/s^2

or if its not negative

27m/s/2.5=10.8m/s^2

3 0

2 years ago