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3 years ago

Based on what you have just read, what are the advantages of storing music digitally compared to using an analog system?

2 answers:

7 0

Storing music digitally requires less storage room than analog records or tapes. Digital music is easier to copy and the copies are the same as the original. The quality of the signal does not degrade over long periods of time.

scZoUnD [109]3 years ago

6 0

The correct answer (sample response) is:

Based on what you have just read the advantages of storing music digitally compared to using an analog system are storing music digitally requires less storage room than analog records or tapes, digital music is easier to copy and the copies are the same as the original. The quality of the signal does not degrade over long periods of time.

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A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

valina [46]

The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
$T=2 t = 2 \cdot 0.100 s = 0.200 s$
Which means that the frequency is
$f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz$
and the angular frequency is
$\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s$

In a spring-mass system, the maximum velocity of the object is given by
$v_{max} = A \omega$
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
$v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s$

6 0

3 years ago

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

where m= mass of the truck = 2.5*10³ kg.
So, we can find the speed v, as follows:

$v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

$W = F*d$

We can solve for F, as follows:

$F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$

4 0

2 years ago

Melanye and Kaitlynn were traveling on a train for 6 hours. Over that time the train was at an average velocity of 70km/hr west.

slega [8]

The way he got the wrong guy to get back into the car he was just saying that I didn’t need it for me so he said no I’m gonna do tell you I have a little girl that was going on a little bit of a girl and I

8 0

3 years ago

An ambulance with a siren emitting a whine at 1790 Hz overtakes and passes a cyclist pedaling a bike at 2.36 m/s. After being pa

Deffense [45]

Answer:

The speed of the ambulance is 4.30 m/s

Explanation:

Given:

Frequency of the ambulance, f = 1790 Hz

Frequency at the cyclist, f' = 1780 Hz

Speed of the cyclist, v₀ = 2.36 m/s

let the velocity of the ambulance be 'vₓ'