All categories

3 years ago

Which best describes the energy of a sound wave as it travels through a medium?

Physics

2 answers:

qaws [65]3 years ago

7 0

Answer:

it increases

Explanation:

earnstyle [38]3 years ago

4 0

Answer:

Im pretty sure is C remains the same not 100% though

Explanation:

Because it is refering to energy not speed

You might be interested in

One strategy in a snowball fight is to throw

faltersainse [42]

Answers:

a) $\theta_{2}=23\°$

b) $t=1.199 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=11.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=67\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(11.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(67\°))$ (9)

$x=9.043 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=45.99\°$

$\theta_{2}=22.99\° \approx 23\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(11.1 m/s)sin(67\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.085 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(11.1 m/s)sin(23\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=0.885 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.085 s - 0.885 s$

Finally:

$t=1.199 s$

7 0

2 years ago

What are precautions in a simple pendulum experiment

monitta

Explanation:

1. draught

2. Parallax error

3. angle if displacement

4. air resistance or any form of obstruction

6 0

3 years ago

What is the resistivity of a wire of 1.3 mm diameter, 2.3 m length, and 61 ms resistance. Number Units The number of significant

kakasveta [241]

Answer:

$\rho = 3.68 *10^{-8} Ω. m$

Explanation:

The resistance is given as

$R = \frac{\rho L}{A}$

Where A IS Cross sectional area of wire $= \pi r^{2}$

therefore resistivity \rho can be wrtten as

$\rho = \frac{\pi Rd^{2}}{4L}$

Putting all value to get resistivity value $\rho =\frac{\pi* 61*10^{-3}* (1.33*10^{-3})^{2}}{4*2.3}$

$\rho = 3.68 *10^{-8} Ω. m$

3 0

3 years ago

An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal

joja [24]

Complete Question

An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?

Answer:

The height is $H = 318.5 \ m$