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3 years ago

What is the difference between three types of mechanical waves?

Physics

2 answers:

Amanda [17]3 years ago

8 0

<u>Answer:</u>

There are three kinds of mechanical waves depending upon the direction in which they travel or propagate through the medium.

(i) Longitudinal waves

(ii) Transverse waves

(iii) Surface waves

In transverse waves, the particles tend to move in a direction at right angles to the waves whereas in longitudinal waves, they move in parallel direction to each other. When it comes to surface waves they lie in circular motion.

Marta_Voda [28]3 years ago

5 0

Answer:

In longitudinal waves, particles travel in the direction parallel to that of the wave motion where as in transverse waves, particles move perpendicular to the direction of the wave motion and in surface waves, particles in the medium move in a circular motion.A good example of longitudinal waves is sound waves. Vibrating a string on the ground can serve as an example of transverse wave.For surface waves, the ocean waves travelling on the surface can illustrate the cicular movement of particles in the water.

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The beat your doctor listens to through a sethoscope is the sound of the four values opening

djyliett [7]

and closing .

The heart has 4 valves. They are what makes the lub-dub lub-dub sounds that can be heard from the chest.

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The tricuspid valve is located between the right atrium and the right ventricle. It closes the right atrium to hold unoxygenated blood and opens to pass it on to the right ventricle ensuring a one way flow.

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3 years ago

An object with a mass of 6.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the amount of the force? R

iren [92.7K]

Answer:

48N

Explanation:

use F=ma, or force is equal to mass multiplied by acceleration.

8 0

3 years ago

Martha wants to calculate an object's velocity. What will she need to do?

rjkz [21]

Answer:

you divide the distance by the time it takes to travel that same distance, then you add your direction to it.

4 0

2 years ago

What is the radius of a tightly wound solenoid of circular cross-section that has 180turns if a change in its internal magnetic

Nata [24]

Answer:

Radius of cross section, r = 0.24 m

Explanation:

It is given that,

Number of turns, N = 180

Change in magnetic field, $\dfrac{dB}{dt}=3\ T/s$

Current, I = 6 A

Resistance of the solenoid, R = 17 ohms

We need to find the radius of the solenoid (r). We know that emf is given by :

$E=N\dfrac{d\phi}{dt}$

$E=N\dfrac{d(BA)}{dt}$

Since, E = IR

$IR=NA\dfrac{dB}{dt}$

$A=\dfrac{IR}{N.\dfrac{dB}{dt}}$

$A=\dfrac{6\ A\times 17\ \Omega}{180\times 3\ T/s}$

$A=0.188\ m^2$

$A=0.19\ m^2$

Area of circular cross section is, $A=\pi r^2$

$r=\sqrt{\dfrac{A}{\pi}}$

$r=\sqrt{\dfrac{0.19}{\pi}}$

r = 0.24 m

So, the radius of a tightly wound solenoid of circular cross-section is 0.24 meters. Hence, this is the required solution.

5 0

3 years ago

Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise fro

marishachu [46]

Answer:

Fe= 28.2 N : Magnitude of the equilibrant (Fe)

β = 18.34° , clockwise from the positive x axis

Explanation:

Concept of the equilibrant

It is called equilibrant to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero

To solve this problem we decompose the forces given into x-y components to find the resulting force:

Look at the attached graphic

F₁= 33.4 N , θ₁=23.8° clockwise from the positive y axis (y+)

F₁x= 33.4 *sin23.8° = 13.48 N

F₁y= 33.4 *cos23.8° =30.6 N

F₂=46.1 N , θ₂=28.8 counterclockwise from the negative x axis (x-)

F₂x= -46.1 *cos28.8° = -40.4 N

F₂y= -46.1 *sin28.8° = -22.2 N

Components of the resultant in x-y R(x,y)

Rx= 13.48 N -40.4 N = - 26.92 N

Ry= 30.6 N -22.2 N = + 8.4 N

Components of the equilibrant in x-y Fe(x,y)

Fex= +26.92 N

Fey= - 8.4 N

Magnitude of the equilibrant (Fe)

$F_{e} = \sqrt{(F_{ex})^{2}+{(F_{ey})^{2} }$

$F_{e} = \sqrt{(26.92)^{2}+(8.4)^{2} }$

Fe= 28.2 N

Angle the equilibrant makes with the x axis ( β)

$\beta = tan^{-1} (\frac{F_{ey} }{F_{ex} } )$

$\beta = tan^{-1} (\frac-8.4 }{26.92 } )$

β = -18.34°

β = 18.34° , clockwise from the positive x axis

8 0

2 years ago