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3 years ago

What is the wavelength of red light in vacuum that has a frequency of 4.00 × 1014 hz?

Physics

1 answer:

Pachacha [2.7K]3 years ago

6 0

Answer: $7.5(10)^{-7} m$

Explanation:

The frequency $\nu$ of an electromagnetic wave is given by the following equation:

$\nu=\frac{c}{\lambda}$

Where:

$c=3(10)^{8} m/s$ is the speed of light in vacuum

$\nu=4(10)^{14} Hz$ is the frequency of red light

$\lambda$ is the wavelength of red light

Isolating $\lambda$ :

$\lambda=\frac{c}{\nu}$

$\lambda=\frac{3(10)^{8} m/s}{4(10)^{14} Hz}$

Finally:

$\lambda=7.5(10)^{-7} m$

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Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging,

Rzqust [24]

A) A. 380 kHz

To clerly see the image of the fetus, the wavelength of the ultrasound must be 1/4 of the size of the fetus, therefore

$\lambda=\frac{1}{4}(1.6 cm)=0.4 cm=0.004 m$

The frequency of a wave is given by

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

where

v is the speed of the wave

$\lambda$ is the wavelength

For the ultrasound wave in this problem, we have

v = 1500 m/s is the wave speed

$\lambda=0.004 m$ is the wavelength

So, the frequency is

$f=\frac{1500 m/s}{0.004 m}=3.75\cdot 10^5 Hz=375 kHz \sim 380 kHz$

B) B. f(c+v)/c−v

The formula for the Doppler effect is:

$f'=\frac{v\pm v_r}{v\pm v_s}f$

where

f' is the apparent frequency

v is the speed of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative if it is moving away from the source)

$v_s$ is the speed of the source (positive if the source is moving away from the receiver, negative if it is moving towards the receiver)

f is the original frequency

In this problem, we have two situations:

- at first, the ultrasound waves reach the blood cells (the receiver) which are moving towards the source with speed

$v_r = +v$ (positive)

- then, the reflected waves is "emitted" by the blood cells (the source) which are moving towards the source with speed

$v_s = -v$

also

v = c = speed of sound in the blood

So the formula becomes

$f'=\frac{c + v}{v - v_s}f$

C. A. The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection

The reflection coefficient is

$R=\frac{(Z_1 -Z_2)^2}{(Z_1+Z_2)^2}$

where Z1 and Z2 are the acoustic impedances of the two mediums, and R represents the fraction of the wave that is reflected back. The acoustic impedance Z is directly proportional to the density of the medium, $\rho$ .

In order for the ultrasound to pass through the skin, Z1 and Z2 must be as close as possible: therefore, a gel with density similar to that of skin is applied, in order to make the two acoustic impedances Z1 and Z2 as close as possible, so that R becomes close to zero.

3 0

3 years ago

An object is 10 cm from the mirror, its height is 1 cm and the focal length is 5 cm. What is the image height? (Indicate the obj

boyakko [2]

The image height is -10 cm (the image is upside down)

Explanation:

We can solve the problem by using the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where:

f = 5 cm is the focal length of the mirror

p = 10 cm is the distance of the object from the mirror

q is the distance of the image from the mirror

Solving for q, we find:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\\\rightarrow q= 10 cm$

So, the distance of the image from the mirror is 10 cm.

Now we can find the image height by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the height of the image

y = 1 cm is the height of the object

and using

p = 10 cm

q = 10 cm

We find the size of the image:

$y' = -\frac{qy}{p}=-\frac{(10)(1)}{10}=-10 cm$

where the negative sign indicates that the image is upside down.

#LearnwithBrainly

8 0

3 years ago

Factors affect the rate of evaporation accept

Debora [2.8K]

Answer: e) turbidity

Explanation:

Evaporation is defined as the process by which a liquid without being heated is converted into vapor.

The factors that affect the rate of evaporation of a liquid are:

a) humidity-When the humidity of an environment is high, it will take more time for water to escape to the atmosphere and less time, if the reverse is the case because Humidity is the amount of water vapor in the atmosphere. For example, on a rainy day, clothes tend to dry slower than a sunny day.

b) nature of liquid- different liquids have different rate of evaporation, For example, alcohol will evaporate faster than water

c) pressure - The higher the pressure on the surface of a body of water, the slower the evaporation rate and vice versa.

d) temperature- The temperature of a surrounding affect how fast a liquid ,(water) can evaporate. When the temperature of the liquid is increased, it raises the kinetic energy of the individual molecules that make up the liquid thereby reducing the inter molecular forces of attraction holding the liquid together and causing fast escape into the atmosphere as a vapor or gas

e) turbidity shows the extent water loses its transparency due to the presence of suspended particles lie clay, silt, organic matter etc and so is not a factor that affect rate of evapouration

7 0

3 years ago

Nathan is walking to the store and sees a snake slithering across the sidewalk. He jumps over it with an initial vertical veloci

Travka [436]

Answer:

$0.62\:\mathrm{s}$

Explanation:

Since the universal SI unit for velocity is meters/second, let's convert ft/s to m/s:

$10\:\mathrm{ft/s}=3.048\:\mathrm{m/s}$

We can use the following kinematics equation to solve this question:

$v_f=v_i+at$

What we know:

The initial velocity, $v_i$ , is $3.048\:\mathrm{m/s}$

(physics concept) The final velocity must be equal in magnitude but opposite in direction to the initial velocity ( $v_f=-3.048\:\mathrm{m/s}$ )
Acceleration, $a$ , is acceleration due to gravity at about $9.8\:\mathrm{m/s}$