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givi [52]
3 years ago
15

Prior to the music CD, stereo systems had a phonographic turntable on which vinyl disk recordings were played. A particular phon

ographic turntable starts from rest and achieves a final constant angular speed of 33 1 3 rpm in a time of 4.5 s.
(a) How many rotations did the turntable undergo during that time?

(b) The classic Beatles album Abbey Road is 47 min and 7 s in duration. If the turntable requires 8 s to come to rest once the album is over, calculate the total number of rotations for the complete start-up, playing, and slow-down of the album.
Physics
1 answer:
mina [271]3 years ago
7 0

Answer: a) 1.3 rev. b) 1,573 rev.

Explanation:

Let's split the process in 3 phases: start-up, playing, and slow-down.

  • Start-Up

As the turntable starts from rest, ω₀ = 0.

If we choose θ₀ = 0º, assuming a constant angular acceleration, we can write the following expression for θ (in rads):

θ = 1/2  γ t² (1)

The angular acceleration γ, by definition is the rate of change of the angular velocity, so:

γ = ωf  - ω₀ / t . (2)

As ω₀ = 0, we can put the following:

γ = ωf / t

As we have ωf in rpm, it is advisable to convert it to rad/sec, as follows:

33 1/3 rpm (1 min/60 sec) . (2π rad/1 rev) = 111/100π rad/sec

Replacing in (2), we have:

γ = 37/150 π rad/sec²

Replacing in (1):

θ₁ = 999/400 π rad = 1.3 rev

  • Playing

During this phase, the turntable rotates at constant angular speed, so we can apply the definition of angular velocity to get the angle rotated, as follows:

θ₂ = ωf . t = (111/100)π . 2,827 sec = 313797/100 π rad = 1,569 rev.

  • Slow-down

Assuming a constant deceleration, as ωf = 0, we can find the value of the angular acceleration during this phase, as follows:

γ = -111/100 π / 8 rad/sec² = -0.44 rad/sec²

Now, we can apply the kinematic equation for the angle rotated under constant angular acceleration condition, as follows:

θ₃ = ω₀ t -1/2γ t²

Replacing by the values, we get:

θ₃ = 222/25 π - 111/25 π = 111/25 π rad = 2.2 rev

Adding the the three angles, we have:

θ = θ₁ + θ₂ + θ₃ = 1.3 + 1,569 + 2.2 = 1,573 rev

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The tension of a guitar string is increased by 40%. By what factor odes the fundamental frequency of vibration change? a. 1.13 b
bogdanovich [222]

Answer:

<h3> b. 1.18</h3>

Explanation:

The fundamental frequency in string is expressed as;

F1 = 1/2L√T/m .... 1

L is the length of the string

T is the tension

m is the mass per unit length

If the tension is increased by 40%, the new tension will be;

T2 = T + 40%T

T2 = T + 0.4T

T2 = 1.4T

The new fundamental frequency will be;

F2 = 1/2L√1.4T/m ..... 2

Divide 1 by 2;

F2/F = (1/2L√1.4T/m)/1/2L√T/m)+

F2/F = √1.4T/m ÷ √T/m

F2/F = √1.4T/√m ×√m/√T

F2/F = √1.4T/√T

F2/F = 1.18√T/√T

F2/F = 1.18

F2 = 1.18F

Hence the fundamental frequency of vibration changes by a factor of 1.18

8 0
3 years ago
How many times does the earth spin in a year
sp2606 [1]
It spins one bc it takes 365 days to make a year and to make a year the earth has to mov some
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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s
Ilia_Sergeevich [38]

Speed of wave is given as

v = 5 m/s

Wavelength of the wave is given as

\lambda = 20 m

now from the formula of wave time period we can say

speed = \frac{wavelength}{time period}

5 = \frac{20}{T}

T = \frac{20}{5}

T = 4 s

so it will have time period of T = 4 s

7 0
3 years ago
Yea, gonna need some help. Thanks
natulia [17]

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t

where,

v₀ = initial speed = 110 ft/s

Therefore,

110 = 32t\\\\t = \frac{110}{32}\\\\

<u>t = 3.48 s</u>

8 0
3 years ago
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