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3 years ago

When a wave is acted upon by an external damping force, what happens to the energy of the wave

Physics

2 answers:

Rasek [7]3 years ago

8 0

The energy of the wave decreases gradually

bazaltina [42]3 years ago

3 0

Answer:

Explanation:

The energy of the wave its determined by the mass of the fuid of the wave and the speed. The amount of motion equation P= m.v

So according to the impulse and amount of motion theorem (ΔP= mvₙ- mv₀).

If the wave is acted upon by an external dumping force, the wave its going to decrease his energy because the reduction of the speed and the mass.

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The Force of friction between an object and the surface upon which it is sliding is 12N. The weight of the object is 20N. What i

VashaNatasha [74]

Answer:

Explanation:

Reducing Sliding Friction. You can reduce the resistive force of sliding friction by applying lubrication between the two surfaces in contact, by using rollers, or by decreasing the normal force

6 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 1.4×109 electrons from one disk to the

allsm [11]

Answer:

r = 6.5*10^-3 m

Explanation:

I'm assuming you meant to ask the diameters of the disk, if so, here's it

Given

Quantity of charge on electron, Q = 1.4*10^9

Electric field strength, e = 1.9*10^5

q = Q * 1.6*10^-19

q = 2.24*10^-10

E = q/ε(0)A, making A the subject of formula, we have

A = q / [E * ε(0)], where

ε(0) = 8.85*10^-12

A = 2.24*10^-10 / (1.9*10^5 * 8.85*10^-12)

A = 2.24*10^-10 / 1.6815*10^-6

A = 1.33*10^-4 m²

Remember A = πr²

1.33*10^-4 = 3.142 * r²

r² = 1.33*10^-4 / 3.142

r² = 4.23*10^-5

r = 6.5*10^-3 m

3 0

3 years ago

Why does light change its velocity (or slows down) when it encounters a glass lens?

Rufina [12.5K]

Explanation:

what happens is that light slows down when it passes from the less dense air into the denser glass or water. This slowing down of the ray of light also causes the ray of light to change direction. It is the change in the speed of the light that causes refraction.

6 0

3 years ago

A woman does 236 J of work

mamaluj [8]

Answer:

Explanation:

Step one:

Given data

work-done in dragging the trash= 236J

applied force= 18.9N

distance moved= 24.4m

Required

The angle of the applied force

Step two:

We know that work done is

WD= F * distance

<em>The work is the product of the horizontal component of the force and the distance. </em>

Horizontal force = 236 ÷ 24.4

= 9.67 N

Cos θ = Horizontal force ÷ Actual force

Cos θ = (236 ÷ 24.4) ÷ 18.9 = 236 ÷ 461.16

The angle is approximately 59˚

6 0

3 years ago