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2 years ago

A wave has a wavelength of 6 mm and a frequency of 9 hertz. What is its speed? what is the answer and units?

1 answer:

5 0

Answer:The speed is 54 mm/s, the units are mm/s.

Explanation:

1) Data:

a) λ = 6mm
b) f = 9 Hz = 9 s⁻¹
c) v = ?

2) Formula:

wave equation: v = f × λ

3) Solution:

v = 9 s⁻¹ × 6mm

v = 54 mm/s

The speed is 54 mm/s, the units are mm/s.

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What is another word for cycles per second??

Hatshy [7]

Frequensey or hertz, I looked this up on the internet!

5 0

2 years ago

Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall,

Genrish500 [490]

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

6 0

2 years ago

A bat emitts a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away

Varvara68 [4.7K]

The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.

From the question given above, the following data were obtained:

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

Time (t) =?

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 × t = 16.84

Divide both side by 343

t = 16.84/343

t = 0.05 s

Thus, the time between when the bat emits the sound and when it hears the echo is 0.05 s.

<h3>How does a bat know how far away something is?</h3>

A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.

Learn more about time elapses between when the bat emits the sound :

<u>brainly.com/question/16931690</u>

#SPJ4

Correction question:

A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)

8 0

1 year ago

All please..............

rodikova [14]

4 water has a low heat capacity and a high vaporization temperate and coasts have low temps

3 0

3 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago