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3 years ago

14

Speed versus Time

1 answer:

Otrada [13]3 years ago

8 0

A since it’s the only one that makes sense! So A is correct.

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You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be

Alex

Answer:

The amplitude is $A = 90.2 \ m$

Explanation:

From the question we are told that

The frequency of when sound is approaching observer is $f = 392 Hz$

The frequency as the move away from observer is $f_ a = 330 \ Hz$

The time between the pitch are $t = 10 \ s$

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

$T = 2 t$

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

$T = \frac{2 \pi}{w}$

Where $w$ is the angular velocity

=> $\frac{2 \pi}{w} = 2 * 10$

=> $w = 0.314 \ rad/sec$

Now using Doppler Effect,

The source of the sound is approaching the observer

The

$f = f_o (\frac{v}{v- wA} )$

$392 = f_o (\frac{v}{v- wA} )$

Where A is the amplitude

So when the source is moving away from the observer

$f_a = f_o (\frac{v}{v+ wA} )$

$330 = f_o (\frac{v}{v+ wA} )$

Here $f_o$ is the fundamental frequency

Dividing the both equation we have

$\frac{392}{330} = \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}$

$1.1878 = \frac{v+wA}{v-wA}$

$1.1878 v - 1.1878 wA = v+wA$

$1.1878 v = 2.1878 wA$

=> $A = \frac{(0.1878 * (330))}{(2.1878)* (0.314)}$

$A = 90.2 \ m$

7 0

3 years ago

Please help ASAP.

Bess [88]

Im not 100% sure you have to tell me if im wrong or not.

D

B

C

3 0

3 years ago

Read 2 more answers

A street light is at the top of a 10 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/sec along a s

Citrus2011 [14]

Answer:

the shadow is moving with 12ft/s from the woman

Explanation:

let the distance from the pole to the shadow tip be L

$\frac{10}{6} = \frac{L}{x}$

$L=\frac{5}{3}x$

$\frac{\mathrm{d} (QR)}{\mathrm{d} t}= 8ft/s$

$\frac{\mathrm{d} (L-x)}{\mathrm{d} t}= 8ft/s$

$\frac{\mathrm{d} (L)}{\mathrm{d} t}-\frac{\mathrm{d} (x)}{\mathrm{d} t}= 8ft/s$

$\frac{\mathrm{d} (L)}{\mathrm{d} t}=\frac{5}{3}\frac{\mathrm{d} (x)}{\mathrm{d} t}$

$\frac{\mathrm{d} (x)}{\mathrm{d} t}=12ft/s$

$\frac{\mathrm{d} (L)}{\mathrm{d} t}=16ft/s$

hence the shadow is moving with 12ft/s from the woman

5 0

3 years ago

A hockey puck with a mass of 0.159 kg slides over the ice. The puck initially slides with a speed of 4.75 m/s, but it comes to a

Anton [14]

Answer:

The energy dissipated as the puck slides over the rough patch is 1.355 J

Explanation:

Given;

mass of the hockey puck, m = 0.159 kg

initial speed of the puck, u = 4.75 m/s

final speed of the puck, v = 2.35 m/s

The energy dissipated as the puck slides over the rough patch is given by;

ΔE = ¹/₂m(v² - u²)

ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)

ΔE = -1.355 J

the lost energy is 1.355 J

Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J

5 0

3 years ago

A(n) blank wave carries energy through space

vova2212 [387]

A transverse wave (electromagnetic wave) carries energy through space.

6 0

3 years ago