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koban [17]
3 years ago
5

What is the exact time of a full earth rotation?

Physics
2 answers:
aniked [119]3 years ago
5 0
The exact time is <span> 23 hours, 56 minutes, and 4 seconds. hope this helps hun</span>
Nina [5.8K]3 years ago
4 0
<span>Partyrockchickk is on the right track, but 23:56:04 is not exactly exact.
23h 56m 4.0906s is closer, but still not exact. You'll never see an 'exact'
number written anywhere. First because it doesn't work out to an exact
decimal of anything, and second because it changes ! The Good Friday
earthquake in Alaska changed it, the Fukushima earthquake changed it,
the Nepal earthquake changed it, and the daily drag of the tides changes
it by something like 0.001 second every century.

One thing's for sure:  It is definitely <em><u>not </u></em>24 hours.


</span>
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A compact disc (CD) stores music in a coded pattern of tiny pits 10⁻⁷m deep. The pits are arranged in a track that spirals outwa
Serggg [28]

a) Angular speed of the innermost part: 50 rad/s, outermost part: 21.6 rad/s

b) The length of the track would be 5,550 m

c) The average angular acceleration is -6.4\cdot 10^{-3}rad/s

Explanation:

a)

For an object in uniform circular motion, the relationship between angular speed and linear speed is

v=\omega r

where

v is the linear speed

\omega is the angular speed

r is the radius of the circle

Here the linear speed of the track is constant:

v = 1.25 m/s

For the innermost part of the disk,

r = 25.0 mm = 0.025 m

So the angular speed is

\omega_i = \frac{v}{r_i}=\frac{1.25}{0.025}=50 rad/s

For the outermost part,

r = 58.0 mm = 0.058 m

So the angular speed is

\omega_o = \frac{v}{r_o}=\frac{1.25}{0.058}=21.6 rad/s

b)

The maximum playing time of the disk is

t=74.0 min \cdot (60 s/min)=4440 s

If the track was stretched out in a straight line, the motion of the track would be a uniform motion (since it is moving at constant speed), so we can use the equation

v=dt

where

v is the linear speed

d is the length coverted in a straight line

t is the time

Substituting v = 1.25 m/s and solving for d,

d=vt=(1.25)(4440)=5550 m

c)

The angular acceleration is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f is the final angular speed

\omega_i is the initial angular speed

t is the time

A CD is read from the innermost part to the outermost part, so we have:

\omega_i = 50 rad/s (at the innermost part)

\omega_f = 21.6 rad/s (at the outermost part)

t = 74.0 min = 4440 s is the time

Substituting, the average angular acceleration is

\alpha = \frac{21.6-50.0}{4440}=-6.4\cdot 10^{-3}rad/s

where the negative sign means the CD is decelerating.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

6 0
3 years ago
What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w
Tatiana [17]

Answer:

a. \ f_1=7.9057Hz\\\\b. \ f_2=15.8114Hz\\\\c. \ f_3=23.7171Hz

Explanation:

a. The wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu} #where t=250N*10=2500N, \mu=0.1kg

#substitute for actual values for the lowest frequency.

F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{1}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=7.9057Hz  #n=1, lowest frequency

Hence, the lowest frequency for standing waves is 7.9057Hz

b.The wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}           #where t=250N*10=2500N,\mu=0.1kg

#The second lowest frequency happens at n=2:

F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{2}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=15.8114Hz

Hence, the second lowest frequency is 15.8114Hz

c.Given that the wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}                    #where t=250N*10=2500N,\mu=0.1kg

The third lowest frequency happens at n=3

F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{3}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=23.7171Hz

Hence, the third lowest frequency is 23.7171Hz

6 0
3 years ago
Which term describes the amount of water vapor in the air?
loris [4]
That would be humidity
4 0
3 years ago
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Find the volume of a box with length 25 cm, height 25 cm and width 1.0 m.
emmasim [6.3K]
Before finding the volume, it is always good to change the units as same.
length 25cm
height 25cm
width 100cm

volume
length * height * width
25 * 25 * 100
62500 cm^3
5 0
3 years ago
what will be the acceleration due to gravity at up planet whose mass is 8 times the mass of the earth and whose radius is twice
Alexxandr [17]
Hope it cleared your doubt.

3 0
3 years ago
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