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3 years ago

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What Does This Poem Mean?

1 answer:

Ann [662]3 years ago

7 0

Nakakakak skskskksksksks

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How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 1.6 % increase

morpeh [17]

Answer:

The value is $v_o = 5.488 \ m/s$

Explanation:

From the question we are told that

The emitted frequency increased by $\Delta f = 1.6 \% = 0.016 \$

Let assume that the initial value of the emitted frequency is

$f = 100\% = 1$

Hence new frequency will be $f_n = 1 + 0.016 = 1.016$

Generally from Doppler shift equation we have that

$f_1 = [\frac{ v \pm v_o}{v \pm + v_s } ] f$

Here v is the speed of sound with value $v = 343 \ m/s$

$v_s$ is the velocity of the sound source which is $v_s = 0 \ m/s$ because it started from rest

$v_o$ is the observer velocity So

Generally given that the observer id moving towards the source, the Doppler frequency becomes

$f_1 = [\frac{ v + v_o}{v + 0 } ] f$

=> $1.016 = [\frac{ 343 + v_o}{343 } ] * 1$

=> $v_o = 5.488 \ m/s$

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3 years ago

What is a real-world application that depends on the relationship between distance, average speed, and time?

sertanlavr [38]

time thats the anwser

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4 years ago

Read 2 more answers

A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s,

Serhud [2]

By using the equation speed = distance/time we can solve for distance. The speed is 4 m/s and the time is 12 seconds. We need to rearrange the equation to Speed * Time = distance. 4(12) = 48; 48 = distance. The cliff is 48 meters high.

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3 years ago

It is recommended that you said at least how many inches away from the airbag ?

Morgarella [4.7K]

10 inches is the correct answer.

4 0

3 years ago

A body on the surface of a planet with the same radius as Earths weighs 10 times more than it does on Earth. What is the mass of

nadya68 [22]

Answer:

Explanation:

Given

radius of Planet is equal to radius of Earth

$r_p=r_e$

Weight of body on Planet $F_p=mg_p$

where m=mass of body

$g_p=acceleration\ due\ to\ gravity\ on\ surface\ of\ Planet$

Weight of body on earth $F_e=mg_e$

$g_e=acceleration\ due\ to\ gravity\ on\ Earth$

acceleration due to gravity is given by

$g=\frac{GM}{r^2}$

where G=gravitational constant

M=mass of Planet

r=radius of planet

for earth $g_e=\frac{GM_e}{r_e^2}$

for planet $g_p=\frac{GM_p}{r_p^2}$

substituting these values in $F_e$ and $F_p$

$F_p=m\times \frac{GM_p}{r_p^2}---1$

$F_e=m\times \frac{GM_e}{r_e^2}---2$

divide 1 and 2

$\frac{F_p}{F_e}=\frac{m\times \frac{GM_p}{r_p^2}}{m\times \frac{GM_e}{r_e^2}}$

$10=\frac{M_p}{M_e}$

$M_p=10M_e$

6 0

3 years ago