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3 years ago

14

A mosquito flaps its wings 600 vibrations per second which produces the annoying 600Hz buzz. The speed of sound is 340 m/s. How

far does the sound travel between wing beats?

1 answer:

erik [133]3 years ago

8 0

Answer:

$\lambda=0.57m$

Explanation:

The distance traveled by the sound wave between the beats of the wing is its wavelength, since it is defined as the distance traveled by a periodic disturbance that propagates in a cycle, is defined as:

$\lambda=\frac{v}{f}$

Where v is the speed of sound and f the frequency of the wing beats

$\lambda=\frac{340\frac{m}{s}}{600Hz}\\\lambda=0.57m$

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2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

1 year ago

Which of the following is an example of static friction?

Tema [17]

The answer is A.......

5 0

3 years ago

Okay okay two questions heh

tensa zangetsu [6.8K]

The answer for question 2 i guess it’s c

3 0

2 years ago

Read 2 more answers

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn

elena-s [515]

Answer:

The change in potential energy is $\Delta PE = - 3.8*10^{-16} \ J$

Explanation:

From the question we are told that

The magnitude of the uniform electric field is $E = 950 \ N/C$

The distance traveled by the electron is $x = 2.50 \ m$

Generally the force on this electron is mathematically represented as

$F = qE$

Where F is the force and q is the charge on the electron which is a constant value of $q = 1.60*10^{-19} \ C$

Thus

$F = 950 * 1.60 **10^{-19}$

$F = 1.52 *10^{-16} \ N$

Generally the work energy theorem can be mathematically represented as

$W = \Delta KE$

Where W is the workdone on the electron by the Electric field and $\Delta KE$ is the change in kinetic energy

Also workdone on the electron can also be represented as

$W = F* x *cos( \theta )$

Where $\theta = 0 ^o$ considering that the movement of the electron is along the x-axis

So

$\Delta KE = F * x cos (0)$

substituting values

$\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)$

$\Delta KE = 3.8*10^{-16} J$

Now From the law of energy conservation

$\Delta PE = - \Delta KE$

Where $\Delta PE$ is the change in potential energy

Thus

$\Delta PE = - 3.8*10^{-16} \ J$

7 0

2 years ago

What is the average speed of a cheetah that runs 88 m in 5 seconds

vovangra [49]

So, the average speed of the Cheetah is 17.6 m/s.

<h3>Introduction</h3>

Hello ! I'm Deva from Brainly Indonesia. This time, I will help regarding the average speed. The average speed is obtained from finding the average of the speeds that occur or can be detected from the division between distance and travel time. The average speed can be formulated by :

$\boxed{\sf{\bold{\overline{v} = \frac{s}{t}}}}$

With the following condition :

$\sf{\overline{v}}$ = average speed (m/s)
s = shift or distance objects from initial movement (m)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

s = shift = 88 m
t = interval of the time = 5 seconds

What was asked :

$\sf{\overline{v}}$ = average speed = ... m/s

Step by step :

$\sf{\overline{v} = \frac{s}{t}}$

$\sf{\overline{v} = \frac{88}{5}}$

$\boxed{\sf{\overline{v} = 17.6 \: m/s}}$

So, the average speed of the Cheetah is 17.6 m/s.

5 0

2 years ago