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3 years ago

Find the wavelength of the ultrasonic wave emitted by a bat if it has a frequency of 4.0 * 10^4 Hz.

Physics

2 answers:

Oliga [24]3 years ago

6 0

ultrasonic wave is a type of sound wave. speed of sound in air at room temp is 343m/s.

speed = wavelength * frequency

wavelength = speed / frequency = 343/4*10^4

= 0.00858m

= 8.58mm

morpeh [17]3 years ago

5 0

Remark

You have to pick a velocity for an ultra sound wave. I'm going to take the sound velocity to be 331 m/s in air, because I'm assuming that is what is meant: it is the sound a bat would make once he emits the ultrasound to a microphone like device that receives it. The media through which that happens is air.

Givens

v = 331m/s

f = 4 *10^4 hz

Formula

velocity = wavelength * frequency

v = $\lambda$ *f

331/(4*10^4) = $\lambda$

0.008275 = $\lambda$

8.275*10^-3 = $\lambda$

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Explain how the potential energy of two charged particles depends on the distance between the charged particles and on the magni

maks197457 [2]

Answer:

According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges

Explanation:

According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges. Since the potential energy of two charged particles is directly proportional to the product of the two charges, its magnitude increases as the charges of the particles increases. For like charges, the potential energy is positive(the product of the two alike charges must be positive) and since potential energy is inversely proportional to the distance between the charges therefore it decreases as the particles get farther apart . For opposite charges, the potential energy is negative(the product of the two opposite charges must be negative) and since potential energy is inversely proportional to the distance between the two charges, it becomes more negative as the particles get closer together.

8 0

3 years ago

A particle is moving along the x-axis so that its position at any time t is greater than and equal to 0 is given by x(t)=2te^-t?

erastovalidia [21]

For speed you can differentiate the equation, for acceleration you can again differentiate the equation .
at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left

6 0

3 years ago

A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p

Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0

2 years ago

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0

3 years ago

This. please help me.

kramer

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2 years ago