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3 years ago

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What is the frequency of sound waves whose wavelength is 0.25 m on a day when the air temperature is 25 degree Celsius? Group of

answer choices
1200 Hz
1384 Hz
1730 Hz
2400 Hz
3000 Hz

2 answers:

anzhelika [568]3 years ago

8 0

Answer:

Frequency, f = 1384 Hz

Explanation:

It is given that,

Wavelength of the sound wave, $\lambda=0.25\ m$

Air temperature, T = 25° C

The speed of sound at a particular temperature is calculated as :

$v=331.5+0.6\times T$

$v=331.5+0.6\times 25$

v = 346.5 m/s

Let f is the frequency of the sound wave. It can be calculated as :

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

$f=\dfrac{346.5\ m/s}{0.25\ m}$

f = 1386 Hz

or

f = 1384 Hz

So, the frequency of sound waves is 1384 Hz. Hence, this is the required solution.

Dvinal [7]3 years ago

5 0

Answer:1384 Hz

Explanation:

Given

wavelength $(\lambda )$ =0.25 m

Temperature T $=25^{\circ}$

at $T=25^{\circ}$ velocity of sound is 346 m/s

and we know

$velocity=frequency\times \lambda$

$v=f\times \lambda$

$346=f\times 0.25$

f=1384 Hz

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2 years ago

A spectral line from a star is observed to have a wavelength of 656.5 nm. The rest wavelength of this line is 656.8 nm. (a) What

nignag [31]

Answer:

speed of star is 1.37 × $10^{5}$ m/s

it is approaching to earth because wavelength of star is decreasing than rest

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

Explanation:

given data

wavelength = 656.5 nm

rest wavelength = 656.8 nm

to find out

the speed of the star , is it approaching

solution

we know here equation that is

wavelength shift / wavelength at rest = velocity of source / speed of light

so put all value and find

velocity of source = 3 × $10^{-9}$ × 3 × $10^{8}$ / 656.8 × $10^{-9}$

velocity of source star = 1.37 × $10^{5}$ m/s

and

it is approaching to earth because wavelength of star is decreasing than rest

and

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3 years ago

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is $8.75 m/s^2$ in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, $F_f = 15 N$ , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, $W=mg$ , where m is the mass of the book and $g=9.8 m/s^2$ is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

$W=(4)(9.8)=39.2 N$

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

$N-W=ma$

where a is the acceleration; however, since N = W, this becomes

$a=0$

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, $F_f = 15 N$ , backward

Since they act along the same line, we can calculate their resultant as

$\sum F = F - F_f = 50 - 15 = 35 N$

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

Here we have:

$\sum F = 35 N$ (in the forward direction)

m = 4 kg

Therefore, the acceleration is

$a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2$ (forward)

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3 years ago

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful

balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

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the angle between B and L is given by:

$\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°$

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

$|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N$

hence, the magnitude of the magnetic force is 0.187N

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3 years ago

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Vlad1618 [11]

Solid is the answer.

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3 years ago