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3 years ago

Select the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz.

1 answer:

6 0

It is defined that a string that is in its two ends, its fundamental frequency and its subsequent frequencies are the product of the whole depending on the number of the harmonic. In other words, harmonics (higher) define the subsequent frequencies under the functions 2f, 3f, 4f, 5f, etc.

Therefore we have that the higher harmonics would be:

1 x 80 Hz = 80 Hz (1st harmonic and Fundalmental Frequency)

2 x 80Hz = 160Hz (2nd harmonic)

3 x 80Hz = 240Hz (3rd harmonic)

4 x 80Hz = 320Hz (4th harmonic)

5 x 80Hz = 400Hz (5th harmonic)

Hence, the frequencies 160Hz and 240Hz are the two higher harmonics of string with a fundamental frequency of 80Hz.

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The mass of Object 2 is double the mass of Object 5. The mass of Object 4 is half of the mass of Object 5 and the mass of Object

SVETLANKA909090 [29]

This is a great problem if you like getting tied up in knots
and making smoke come out of your brain.

I found that it makes the problem a lot easier if I give the objects some
numbers. I'm going to say that the mass of Object 5 is 20 clods.

Let the mass of Mass of Object 5 be 20 clods .

Then . . .

-- The mass of Object 2 is double the mass of Object 5 = 40 clods.

-- The mass of Object 4 is half of the mass of Object 5 = 10 clods.
and
-- the mass of Object 3 is half of the mass of Object 4 = 5 clods.

So now, here are the masses:

Object #1 . . . . . unknown
Object #2 . . . . . 40 clods
Object #3 . . . . . 5 clods
Object #4 . . . . . 10 clods
Object #5 . . . . . 20 clods .

Now let's check out the statements, and see how they stack up:

Choice-A:
Object 3 and Object 5 exert the same gravitational force on Object 1.
Can't be.
Objects #3 and #5 have different masses, so they can't both
exert the same force on the same mass.

Choice-B.
Object 2 and Object 4 exert the same gravitational force on Object 1.
Can't be.
Objects #2 and #4 have different masses, so they can't both
exert the same force on the same mass.

Choice-C.
The gravitational force between Object 1 and Object 2 is greater than
the gravitational force between Object 1 and Object 4.
Yes ! Yay !
Object-2 has more mass than Object-4 has, so it must exert more force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Choice-D.
The gravitational force between Object 1 and Object 3 is greater than the gravitational force between Object 1 and Object 5.
Can't be.
Object-3 has less mass than Object-5 has, so it must exert less force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Conclusion:
If the DISTANCE is the same for all the tests, then Choice-C is
the only one that can be true.

8 0

3 years ago

A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha

dlinn [17]

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

$a=(-2v^3)\ m/s^2$

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=-2v^3$

$\dfrac{dv}{-v^3}=2dt$

On integrating

$int{\dfrac{dv}{-v^3}}=\int{2dt}$

$\dfrac{1}{2v^2}=2t+C$

$v^2=\dfrac{1}{4t+2C}$ ....(I)

At t = 0, v = 10 m/s

$10^2=\dfrac{1}{4\times0+2C}$

$C=\dfrac{1}{200}$

Put the value of C in equation (I)

$v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}$

$v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}$

$v=0.099\ m/s$

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

$\dfrac{ds}{dt}=v$

On integrating

$\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt$

$s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'$

At t = 0, s = 15 m

$15=\dfrac{200}{800}+C'$

$C'=15-\dfrac{200}{800}$

$C'=14.75$

Put the value in the equation

$s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75$

$s=19.75\ m$

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

3 0

3 years ago

What happened 1 billion years after the Big Bang?

maks197457 [2]

Explanation:

1 billion years after the big bang, the temperature is 20K and some stars and galaxies began to contract due to the gravitational contraction of the over densities of the previous universe. At about 10 billion years after the big bang, our earth and sun form

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4 years ago

if you went to the doctor 3 weeks ago && youre going back, will they have to measure your height and weigh you again?

tino4ka555 [31]

No, they usually will not. It also depends why you're going/why you went the first time.

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3 years ago

Read 2 more answers

what is the magnitude of the compression forces (assumed to be horizontal) acting on both sides of the center board that is sand

Len [333]

F = normal force by each board on each side

W = weight of the board in between acting in down direction = 95.5 N