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2 years ago

In which applications are sound waves reflected? Check all that apply.

Physics

2 answers:

mojhsa [17]2 years ago

8 0

detecting diseases, locating damaged cells, monitoring fetal development

DanielleElmas [232]2 years ago

3 0

Answer: The correct answers are detecting diseases, locating damaged parts and monitoring fetal development.

Explanation:

Ultrasound is a sound wave with higher frequency. The frequency of ultrasound wave is higher than 20,000 Hz.

Ultrasound is a technology that uses high frequency sound waves to create images.

Sound wave is used to detect the damaged parts in the big machinery plants by sending the sound waves. If the sound wave gets reflected back then there is no flaws in the machine but it is not then there may be some damaged parts or cracks in that machine.

Ultrasound is used in medical imaging technique. The ultrasound machine transmits high frequency sound into body by using probe to get the images of the internal parts of the body. It helps in monitoring fetal development. It detects the diseases in the body like tumor.

Therefore, the applications of the sound waves reflected are detecting diseases, locating damaged parts and monitoring fetal development.

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Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass. Compute the

fenix001 [56]

Answer:

Answered

Explanation:

a) Two balls are at a distance of L/2 from the axis of rotation and one block at the center. ( center of rod).

therefore,

$I=2\times m\frac{L}{2}^2$

$I= \frac{1}{2}mL^2$

b) two balls at a distance L/4 at the from the axis and 1 ball at a distance 3L/4 from the from the axis.

$I= 2\times m\times(L/4)^2 + m(\frac{3L}{4})^2$

= $\frac{1}{16} mL^2(2+9)= \frac{11}{16}mL^2$

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2 years ago

"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati

Rasek [7]

Answer:

99.95%

Explanation:

A double pulsar system named PSR J0737-3039A/B in Puppis constellation was discovered in the year 2003. Pulsars are second most densest object in the universe after black holes and they emit radio waves at regular intervals. This pair presented a great and natural setup to test the Theory of General Relativity presented by Einstein in 1915. In this theory Einstein had presented a set of equations on how the space-time fabric will be curved because of the very dense objects such as Neutron stars. It also predicted how the gravitational waves are created because of stars orbiting each other.

A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.

8 0

2 years ago

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli

Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: $sin \theta=\frac{m\lambda}{\alpha}$

Given $m=3$ , $\lambda=6.33\times 10^-^7$ and $\alpha=0.00015$ :

#Substitute our variable values in the minima equation to obtain $\theta$ :

$\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad$

#draw a triangle to find the relationship between $\theta, y \$ and $L$ .

$tan(\theta)=y/L$ #where $y=4.05cm$

$L=y/tan(\theta)=3.20$

Hence the screen is 3.20m from the split.

b. To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

$sin \theta=\frac{m\lambda}{\alpha}$ , where $m=4$ given that it's the minima with the smallest distance.

$sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad$

#we then use $tan(\theta)=y/L$ to calculate $L$ =4.5cm

Then from the equation subtract $y_3$ from $y$ :

$4.50cm-4.05cm=0.45cm$

Hence, the distance $\bigtriangleup y$ is 0.45cm

8 0

2 years ago

Physics in motion unit 6a the nature of waves

mylen [45]

What’s the question?

7 0

2 years ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

2 years ago