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3 years ago

9

-g A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the

motion is 0.165 m. The maximum speed of the block is 3.90 m/s. What is the maximum magnitude of the acceleration of the block?

1 answer:

Nina [5.8K]3 years ago

5 0

Answer:

a= 92. 13 m/s²

Explanation:

Given that

Amplitude ,A= 0.165 m

The maximum speed ,V(max) = 3.9 m/s

We know that maximum velocity in the SHM given as

V(max) = ω A

ω=Angular speed

A=Amplitude

$\omega =\dfrac{3.9}{0.165}\ rad/s$

ω=23.63 rad/s

The maximum acceleration given as

a = ω² A

a= (23.63)² x 0.165 m/s²

a= 92. 13 m/s²

Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².

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The fundamental frequency of a guitar string is 367 hz . part a what is the fundamental frequency if the tension in the string i

ANTONII [103]

The fundamental frequency of a string is given by:
$f_1 = \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where L is the string's length, T the tension and $\mu$ the linear density of the string.

We can see that f1 is proportional to the square root of T: $\sqrt{T}$ .
This means that if the new tension is half the initial value, the new fundamental frequency will be proportional to $\sqrt{ \frac{T}{2} }= \frac{ \sqrt{T} }{ \sqrt{2} }= \frac{f_1}{ \sqrt{2} }$

So, the new fundamental frequency will be
$f_1 ' = \frac{367 Hz}{ \sqrt{2} }=259.5 Hz$

6 0

4 years ago

You must exert a force of 5N on a book to slide it across a table.If you do 2.5 J of work in the process, how far has the book m

I am Lyosha [343]

Answer:

0.5 m.

Explanation:

From the question given above, the following data were obtained:

Force (F) = 5 N

Work done (W) = 2.5 J

Distance (s) =?

Workdone is simply defined as the product of force (F) and distance (s) moved in the direction of the force. Mathematically, it is expressed as:

Work done (W) = Force (F) × Distance (s)

W = F × s

With the above formula, we can obtain the distance to which the book has moved as follow:

Force (F) = 5 N

Work done (W) = 2.5 J

Distance (s) =?

W = F × s

2.5 = 5 × s

Divide both side by 5

s = 2.5/5

s = 0.5 m

Therefore, the book moved 0.5 m when the force was applied.

3 0

4 years ago

Two identical traveling waves, moving in the same direction, are out of phase by π/5.0 rad. What is the amplitude of the resulta

andreev551 [17]

Answer:

Therefore the amplitude of the resultant wave is $=0.95 y_m$

Explanation:

The equation of wave:

y=A sin (kx-ωt)

For wave 1:

y₁=A sin (kx-ωt) = $y_{m}$ sin (kx-ωt)

For wave 2:

y₂=A sin (kx-ωt+Φ) = $y_{m}$ sin (kx-ωt+Φ)

Where A= amplitude= $y_m$

The angular frequency $\omega=\frac{2\pi}{T}$

$k=\frac{2\pi}{\lambda}$ , $\lambda$ = wave length.

t= time

T= Time period

$\phi$ = phase difference = $\frac{\pi}{5}$

The resultant wave will be

y = y₁ + y₂

= $y_m$ sin (kx-ωt) + $y_m$ sin (kx-ωt+Φ)

$=y_m$ {sin (kx-ωt) + sin (kx-ωt+Φ)}

$=y_m\ sin(\frac{kx-\omega t +\phi + kx-\omega t }2)\ cos(\frac{kx-\omega t +\phi -kx+\omega t}2)$

$=y_m\ sin({kx-\omega t +\frac\phi 2)\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\phi }2) sin({kx-\omega t +\frac\phi 2)$

Therefore the amplitude of the resultant wave is

$=y_m\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\pi }{10})$

$=0.95 y_m$

6 0

4 years ago

A slide whistle is an open-closed tube with an adjustable plunger that changes the length. You are playing the slide whistle and

Serjik [45]

Answer:

df / ft = -n 12 n= 1, 3, 5, ...

Explanation:

The speed of sound is

v = λ f

In a whistle that we approach by an open tube at one end and closed at the other, standing waves occur, which has a node in the closed part and a maximum in the open pate, whereby wavelength and the distance of the tube are related, the fundamental wave is

λ₁ = 4L

The harmonics are

λ₃ = 4L / 3

λ₅ = 4L / 5

The general formula

λₙ = 4L / n

with n = 1, 3, 5,…

We substitute and clear in the first equation

f = v n / 4L n = 1, 3, 5,…

Let's use derivatives to find the frequency change

df / dt = v n /4 dL⁻¹ / dt

d / dt (1/L) = - 1 / L² dL / dt

Where dL / dt = 8 cm / s

We replace

df / dt = - n v / L2 dL / dt

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4 0

4 years ago

48 You are given two amplifiers, A and B, to connect in cascade between a 10-mV, 100-k source and a 100- load. The amplifiers ha

Sedbober [7]

Answer:

SABL

Explanation:

The best amplifier will be the one that gives us a bigger gain. In each stage will be a load factor that will reduce the gain, that is defined as:

$Fp=\frac{R_{in}}{R_{in}+R_{out}}\\$

where Rin is the input resistance of the next stage and Rout the output resistance of the previous stage.

Analyzing SABL:

$Fp_1=\frac{100K}{100K+100K}=0.5\\\\Fp_2=\frac{10K}{10K+10K}=0.5\\\\Fp_3=\frac{100}{100+1K}=0.0909$

the total gain will be the total gain of each stage multiplied by the load factor.

$SABL_{gain}=0.5*100*0.5*10*0.0909\\SABL_{gain}=22.73\\SABL_{gain_{db}}=20*\log{22.73}=27.13db$

Analyzing SBAL:

$Fp_1=\frac{10K}{10K+100K}=0.0909\\\\Fp_2=\frac{100K}{100K+10K}=0.909\\\\Fp_3=\frac{100}{100+10K}=0.0099$

the total gain will be the total gain of each stage multiplied by the load factor.

$SABL_{gain}=0.0909*10*0.99*100*0.0099\\SABL_{gain}=0.89\\SABL_{gain_{db}}=20*\log{0.89}=-1.01db$

So the best amplifier arrangement is SABL.

4 0

3 years ago