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3 years ago

Which type of wave does the illustration depict?

Physics

2 answers:

NeX [460]3 years ago

5 0

Answer is B. Longitudinal Wave

Veronika [31]3 years ago

3 0

Answer : The correct option is, (b) longitudinal wave

Explanation :

Longitudinal waves : It is defined as the waves in which the particles of the medium move in the direction of the wave.

Transverse wave : It is defined as the waves in which the particles of the medium travel perpendicularly to the direction of the wave.

Surface wave : It is defined as a combination of transverse and longitudinal waves.

From the given image we conclude that, this illustration depict the longitudinal wave because the particles of the medium move in the direction of the wave.

Hence, the correct option is, (b) longitudinal wave

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A person skateboarding is part of a closed system that has 1,450 j of

soldier1979 [14.2K]

The gravitational potential energy of the system will decreases from 1,250 J to 625 J. Option A is corect.

<h3>What is the law of conservation of energy?</h3>

According to the Law of Conservation of Energy, energy can neither be created nor destroyed, but it can be transferred from one form to another.

The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

TE=KE+GPE

Case 1;

1450 = 200 J+GPE

GPE=1450 -200

GPE=1250 J

Case 2;

1450 = 825 J+GPE

GPE=1450 -825

GPE=625 J

The gravitational potential energy of the system will decreases from 1,250 J to 625 J.

Hence, option A is corect.

To learn more about the law of conservation of energy, refer to brainly.com/question/2137260.

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2 years ago

The use of a laser beam to seal leaky blood vessels and to prevent the growth of new ones in diabetic retinopathy is called lase

emmainna [20.7K]

The answer is photocoagulation.

The use of a laser beam to seal leaky blood vessels and to prevent the growth of new ones in diabetic retinopathy is called laser <u>photocoagulation.</u>

<u></u>

What is photocoagulation?

A minimally invasive method used to treat numerous retinal illnesses is photocoagulation of the retina, also known as retinal laser photocoagulation. The retina may expand due to aberrant leaky blood vessels developing across it in a number of disorders. Laser photocoagulation uses thermal energy above 65 °C to burn the retinal tissue by creating thermal burns. This can prevent the retina from being damaged by the bleeding blood vessels. In addition to causing fibrosis, laser photocoagulation can also seal retinal tears. Laser photocoagulation is typically unable to recover already lost vision in cases of retinal disease, but it can slow the progression of the condition, lower the chance of further vision loss, and preserve residual vision. The likelihood of problems following the operation is quite minimal.

To learn more about photocoagulation click on the link below:

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6 0

1 year ago

Read 2 more answers

an asteroid whose mass is 2.0 x 10^-4 times the mass of earth revolves in a circular orbit around the sun at a distance that is

Xelga [282]

Answer:

0.0002

Explanation:

First you have to solve exponents first, so you solve 10 to the power of negative 4 which is 0.0001 and now you have 2.0x0.0001 and that equals 0.0002

5 0

3 years ago

Any change in an objects speed and /or direction ?

Alexeev081 [22]

Answer:

The answer to this question is acceleration or as scientist say velocity.

Explanation: We know that when you drive a car the acceleration is effected by the amount of speed you add when you press the gas,so you are accelerating the car faster.

8 0

2 years ago

5) Consider pushing a 50.0 kg box through a 5.00 m displacement on both a flat surface and up a

Svetach [21]

a) The work done by the gravitational force on the flat surface is zero

b) The work done by the gravitational force on the ramp is -634 J

c) The work done by the applied force on the flat surface is 500 J

d) The work done by the applied force on up along the ramp is 500 J

Explanation:

The work done by a force is given by the equation

$W=Fdcos \theta$

where

F is the magnitude of the force

d is the dispalcement of the object

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we want to calculate the work done by the gravitational force as the box is pushed across the flat ground.

We immediately notice that the gravitational force acts downward, while the displacement is horizontal: therefore, the angle between force and displacement is $90^{\circ}$ ; this means that $cos 90^{\circ}=0$ , and therefore, the work done is zero:

$W=0$

In this case, the box is pushed along the ramp. We have:

$F=mg=(50.0)(9.8)=490 N$ is the magnitude of the force of gravity, where

m = 50.0 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration of gravity

d = 5.00 m is the displacement of the box along the ramp

The ramp is inclined to the horizontal by $15.0^{\circ}$ , therefore the angle between the force of gravity and the displacement of the box (moving up along the ramp) is:

$\theta=90^{\circ}+15^{\circ}=105^{\circ}$

Therefore, the work done by gravity in this case is:

$W=(490)(5.00)(cos 105^{\circ})=-634 J$

In this case, we want to calculate the work done by the force you apply as the box is pushed across the flat ground.

Here we have:

F = 100.0 N (force applied)

d = 5.00 m (displacement of the box)

$\theta=0^{\circ}$ (the force is applied parallel to the flat surface, therefore force and displacement have same direction)

Therefore, the work done by the force you apply on the flat ground is:

$W=(100.0)(5.00)(cos 0^{\circ})=500 J$

In this last case, we want to calculate the work done by the force you apply as the box is pushed up along the ramp.

This time we have:

F = 100.0 N (force applied is the same)

d = 5.00 m (displacement of the box is also the same)

$\theta=0^{\circ}$ (the force is applied parallel to the ramp, therefore force and displacement have again same direction)

Therefore, the work done by the force you apply while pushing the box along the ramp is:

$W=(100.0)(5.00)(cos 0^{\circ})=500 J$

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8 0

3 years ago