All categories

4 years ago

What is the difference between pulse and periodic waves, and what is the difference between transverse and longitudinal waves?

Physics

1 answer:

stiv31 [10]4 years ago

7 0

Difference between pulse and periodic waves:

A pulse wave is a sudden disturbance in which only one wave or a few waves are generated, such as in the example of the pebble. Thunder and explosions also create pulse waves. A periodic wave repeats the same oscillation for several cycles, such as in the case of the wave pool, and is associated with simple harmonic motion. Each particle in the medium experiences simple harmonic motion in periodic waves by moving back and forth periodically through the same positions.

Difference between longitudinal and transverse waves:

A transverse wave propagates when the disturbance is perpendicular to the propagation direction. An example of a transverse wave is where a woman moves a toy spring up and down, generating waves that propagate away from herself in the horizontal direction while disturbing the toy spring in the vertical direction.

In a longitudinal wave, the disturbance is parallel to the propagation direction. Example of longitudinal wave is where the woman now makes a disturbance in the horizontal direction—which is the same direction as the wave propagation—by stretching and then compressing the toy spring.

You might be interested in

An object weighs 100 newtons on earth.What is its weight on the moon?

Pachacha [2.7K]

Answer:

a) The gravitational acceleration at the surface of the Moon is g moon=1.67 m/s

The ratio of weights (for a given mass ) is the ratio of g-values, so

moon

=(100N)(1.67/9.8)=17N.

(b) For the force on that object caused by Earth's gravity to equal 17 N, then the free fall acceleration at its location must be

=1.67m/s

. Thus , .

= r 2

⇒

=1.5×10

So the object would need to be a distance of r/R

=2.4 "radii" from Earth's center.

4 0

3 years ago

Read 2 more answers

When the body requires an increased blood flow rate in a particular organ or muscle, it can accomplish this by increasing the di

horsena [70]

Answer: The percentage increase in diameter is 19%

Explanation:

Using Poiseuille's law which states that Flow rate , Q = ΔPπr⁴/8ηl

where ΔP is pressure difference, π is a constant, r is the radius which is half of the diameter, η is viscosity of blood, l is length of blood vessel

Let Q₁ be the intitial flow rate, Q₂ the final flow rate,r₁ and r₂ the initial and final radius.

Note: r = d/2, therefore, r₁⁴ = d₁⁴/16 r₂⁴ = d₂⁴/16

Also, Q₂ = 2Q₁, since the flow rate is doubled

Since all factors except diameter is constant, d₂⁴/16 = 2d₁⁴/16

d₂⁴/16 = d₁⁴/8

multiply both sides of the equation by 16

d₂⁴ = 2d₁⁴

taking the fourth root of both sides

d₂ = 1.19*d₁

d₂ - d₁ = 1.19d₁ - d₁

d₂ - d₁ = d₁(1.19 - 1)

d₂ - d₁ = 0.19d₁

d₂ - d₁/d₁ = (0.19d₁/d₁)*100%

(d₂ - d₁/d₁)*100% = 19%

Therefore, the percentage increase in diameter is 19%

3 0

4 years ago

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

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

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

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

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago

7. En la sala de una casa hay una gran ventana de vidrio, por la que se presenta una pérdida significativa de calor; las medidas

jonny [76]

Answer:

I DON'T SPEAK TACO BELL!

Explanation:

7 0

3 years ago

A 12kg cheetah accelerates 24 m/s". What is the force the cheetah needed to run?

Kobotan [32]

Answer:

288N

Explanation:

Given parameters:

Mass of Cheetah = 12kg

Acceleration = 24m/s²

Unknown:

Force needed by the cheetah to run = ?

Solution:

The force needed by the Cheetah to run is the net force.

According to Newton's law;

Force = mass x acceleration

Insert the given parameters and solve;

Force = 12 x 24 = 288N

7 0

3 years ago