What is the speed of sound in air with temperature of 355.8 k

A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

$\Omega = \frac{Mgd}{I\omega}$

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

$\omega$ = Angular velocity

Replacing the value for moment of inertia

$\Omega= \frac{MgR}{MR^2 \omega}$

$\Omega = \frac{g}{R\omega}$

The value for our angular velocity is not in SI, then

$\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})$

$\omega = 104.7rad/s$

Replacing our values we have that

$\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}$

$\Omega = 1.17rad/s$

The precession frequency is

$\Omega = \frac{2\pi rad}{T}$

$T = \frac{2\pi rad}{\Omega}$

$T = \frac{2\pi}{1.17}$

$T = 5.4 s$

Therefore the precession period is 5.4s

7 0

2 years ago

Increase in Space Suit Pressure 0.0/3.0 points (graded) If the pressure in a space suit increases, how will each of the followin

Art [367]

Answer:

Flexibility Increases

Pre-breathe time decreases

Mass of suit decreases.

Explanation:

Spacesuits are designed for space shuttles when a person goes to explore the galaxy. The spacesuits shuttle era are pressurized at 4.3 pounds per inch. The gas in the suit is 100% of oxygen and there is more oxygen to breathe when the altitude of 10,000 is reached. This will decrease the breathing time and mass of suit.

3 0

3 years ago

Which of these descriptions best defines the solar system?

Marrrta [24]

Answer:

d

Explanation:

d because the sun is in the center and everything else surrounds it

5 0

2 years ago

The same wave travels for 2 seconds, what is its frequency?*

sveticcg [70]

Answer:

In waves distance is measured by wave length and time is measured by frequency or period.

velocity ratio=wave length multiply by frequency.

HENCE, if the same wave travels for 2 econds its frequency will be 2Hz.

Explanation:

8 0

2 years ago

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

Lisa [10]

Answer:

1) v₀x = 13.76 m/s

2) v₀y = 19.66 m/s

3) ymax = 21.199 m

4) X = 55.1746 m

5) and 6) y = 18.4 m

Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m

4) We can use this equation

y = y₀ + (tan α)*x – (g / (2* v₀x²))*x²

where y = y₀ = 1.5 m

then

1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²

⇒ 0.02588 x² - 1.42815 x = 0

Solving this equation we get

x₁ = 0 and x₂ = 55.1746 m

The distance between the two girls is 55.1746 m

5) and 6) If v₀x = 15 m/s = vx and ymax = 24 m

y = ? when x = (xmax/2)

ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))

⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s

then we get α' as follows

α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s

Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

⇒ y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))

⇒ y = 18.4 m

7 0

3 years ago

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What is the speed of sound in air with temperature of 355.8 k​

What is the speed of sound in air with temperature of 355.8 k