Answer:
71.4583 Hz
67.9064 N
Explanation:
L = Length of tube = 1.2 m
l = Length of wire = 0.35 m
m = Mass of wire = 9.5 g
v = Speed of sound in air = 343 m/s
The fundamental frequency of the tube (closed at one end) is given by
The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz
The linear density of the wire is
The fundamental frequency of the wire is given by
The tension in the wire is 67.9064 N
TheThe distance they have covered for trip is 219Km.
<h3>What do you mean by uniformly accelereted motion?</h3>
When an object is traveling in a straight line with an increase in velocity at equal intervals of time.
The total time for the trip is
T→ t1+ 22 min = t1+ 0.367 h ,where t1 is the time spent traveling at
V1= 89.5 km/ h .
the distance traveled is ∆x = V1t1=Vavg T
after applying value and calculating it we get
t1= 2.44h for a total time of
t total we get 2.81 h .
∆x =V1T1 = VavgTtotal
∆x = 77×2.81= 219 Km
to learn more about Uniform accelereted motion click here brainly.com/question/12920060
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Answer:
the answers the correct one is cη
Explanation:
In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so
sin θ = θ
with this approach the equation will be surveyed
d² θ / dt² = - g / L sin θ
It is reduced to
d² θ / dt² = - g / L θ
in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%
The angle is related to the height of the pendulum
sin θ = h / L
h = L sin θ.
Therefore the student must be careful that the height is small.
When reviewing the answers the correct one is cη
Answer:
AM broadcasts occur on North American airwaves in the medium wave frequency range of 525 to 1705 kHz (known as the “standard broadcast band”). The band was expanded in the 1990s by adding nine channels from 1605 to 1705 kHz.
