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ASHA 777 [7]
3 years ago
5

Parasaurolophus was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.5-m-long hollow tube in the cr

est had connections to the nose and throat, leading some investigators to hypothesize that the tube was a resonant chamber for vocalization. If you model the tube as an open-closed system, what are the first three resonant frequencies?
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
5 0

Answer:

f1 = 58.3Hz, f2 = 175Hz, f3 = 291.6Hz

Explanation:

lets assume speed of sound is 350 m/s.

frequencies of a standing wave modes of an open-close tube of length L

fm = m(v/4L)

where m is 1,3,5,7......

and fm = mf1

where f1 = fundamental frequency

so therefore: f1 = 350 x 4 / 1.5

f1 = 58.3Hz

f2 = 3 x 58.3

f2 = 175Hz

f3 = 5 x 58.3

f3 = 291.6Hz

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A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other
serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, \omega_{rotation} = Angle turned/Time

∴ \omega_{rotation}  = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, \omega _{Star}, is given as follows;

\omega _{Orbit}  = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s.

3 0
3 years ago
A plane moves at 15 km/h for a total distance of 300km. How long did it fly for? *
strojnjashka [21]
20 hours you do 300 divided by 15 and get 20 hours
5 0
3 years ago
Formula one racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance.
soldi70 [24.7K]

Answer:

a = -36.8 m/s/s

Explanation:

Initial speed of the car

v_i = 90 m/s

finally car will stop after it cover the distance

d = 110 m

so we have

v_f^2 - v_i^2 = 2 a d

here we have

0 - 90^2 = 2(a) (110)

a = \frac{90^2}{220}

a = -36.8 m/s^2

8 0
3 years ago
Particle A has half the mass and eight times the kinetic energy of particle B. What is the speed ratio vA / vB?
Marta_Voda [28]

Answer:

Use the equation "KE=½mv²", and use some algebra. > "Particle A has two times the mass...of particle B" mA = 2mB > "Particle A has...8 situations the kinetic power of particle B" KE_A = 8(KE_B) or: ½(mA)(vA)² = 8(½(mB)(vB)²) the rest is uncomplicated algebra: in basic terms sparkling up the above equation for "vA/vB". (hint: start up by utilising dividing the two factors by utilising "(mB)(vB)²". Then make the substitution: mA/mB = 2 (from the 1st eq0.5 ma *  VA^2 = 8 * 0.5 * mb * VB^2

0.5 * 0.5*mb *  VA^2 = 8 * 0.5 * mb * VB^2

0.5 * 0.5*  VA^2 = 8 * 0.5 * VB^2

VA/VB = 4uation))

Explanation:

8 0
3 years ago
a 2.9-cm-diameter parallel-plate capacitor has a 2.0 mm spacing. the electric field strength inside the capacitor is
Elenna [48]
The electric field strength between the plates is<br /><br />(voltage between the plates) / (0.002) <br /><br />volts per meter.
5 0
3 years ago
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