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BlackZzzverrR [31]
2 years ago
12

The amplitude of a wave decreases gradually as the wave travels down a long, stretched string. What happens to the energy of the

wave when this happens? (The frequency is the same).
Physics
1 answer:
Setler79 [48]2 years ago
8 0

Answer:

energy decreases

Explanation:

The energy of a wave is directly proportional to the square of amplitude of a wave.

Amplitude is defined as the maximum displacement of the wave particle from its mean position.

So, as the amplitude goes on decreasing, the energy of the wave is also decreasing.

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A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed
Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

F = \dfrac{mv^2}{r}

m is the mass, v is the velocity and r is the radius.

It follows that F \propto v^2, provided m and r are constant.

When v is doubled, the new force, F_1, is

F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F

Hence, the tension in the string is quadrupled.

8 0
2 years ago
An object moving at a velocity of 32m/s slows to a stop in 4 seconds. What was its acceleration?
Romashka [77]

Answer:

8m/s

Explanation:

a=d/t

a=32/4

a=8 m/s

6 0
2 years ago
Read 2 more answers
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
Determine the torque applied to the shaft of a car that transmits 225 hp
Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

As

T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

So

T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu

8 0
2 years ago
For nitrogen feel like with its temperature must be within 12.78 Fahrenheit of -333.22 Fahrenheit which equation can be used to
photoshop1234 [79]

Answer:

The following equation can be used.

(32°F − 32) × 5/9=C

7 0
2 years ago
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