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swat32
3 months ago
10

In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

he axis of rotation. The cables swing out at a constant angle of 45.0° when the ride is operating. What is the angular speed of rotation?
Physics
1 answer:
ValentinkaMS [3.4K]3 months ago
7 0

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

T\cos\theta=mg

T=\dfrac{mg}{\cos\theta}

Vertical component

T\sin\theta=m\omega^2 r

Substituting the tension value

mg\tan\theta=m\omega^2(d+L\sin\theta)

\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}

Substituting the value into the equation

\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}

\omega=1.34\ rad/s

Thus, the angular speed of rotation computes to 1.34 rad/s.

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The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu
Yuliya22 [3333]

Answer: 0.10233nm

Explanation:

The mean free path of an atom can be calculated using the following equation:

\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}    (1)

Where:

\lambda=0.2\mu m=0.2(10)^{-6}m

R=8.3145J/mol.K is referred to as the Universal gas constant

T=0\°C=273.115K represents the absolute standard temperature

d denotes the diameter of helium atoms

N_{A}=6.0221(10)^{23}/mol symbolizes Avogadro's number

P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3} indicates absolute standard pressure

<pFrom this, we can solve for d using (1), aiming to determine the radius r of the helium atom:

d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}    (2)

d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}    (3)

d=2.0467(10)^{-10}m    (4)

If the radius equals half of that diameter:

r=\frac{d}{2}  (5)

Eventually:

r=\frac{2.0467(10)^{-10}m}{2}  (6)

r=1.0233(10)^{-10}m  (7)

Nonetheless, we were tasked with finding this radius in nanometers. Knowing 1nm=(10)^{-9}m:

r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm  (8)

Ultimately:

r=0.10233nm Represents the radius of the helium atom in nanometers.

5 0
3 months ago
The G string on a guitar is 69 cm long and has a fundamental frequency of 196 Hz. A guitarist can play different notes by pushin
Keith_Richards [3271]
Δd = 23 cm. When the eta string of the guitar has nodes at both ends, the resulting waves create a standing wave, which can be expressed with the following formulas: Fundamental: L = ½ λ, 1st harmonic: L = 2 ( λ / 2), 2nd harmonic: L = 3 ( λ / 2), Harmonic n: L = n λ / 2, where n is an integer. The rope's speed can be calculated using the formula v = λ f. This speed remains constant based on the tension and linear density of the rope. Now, let's determine the speed with the provided data: v = 0.69 × 196, yielding v = 135.24 m/s. Next, we will find the wavelengths for the two frequencies: λ₁ = v / f₁, which gives λ₁ = 135.24 / 233.08, equaling λ₁ = 0.58022 m; λ₂ = v / f₂ results in λ₂ = 135.24 / 246.94, consequently λ₂ = 0.54766 m. We'll substitute into the resonance equation Lₙ = n λ/2. At the third fret, m = 3, therefore L₃ = 3 × 0.58022 / 2, resulting in L₃ = 0.87033 m. For the fourth fret, m = 4, which gives L₄ = 4 × 0.54766 / 2, equating to L₄ = 1.09532 m. The distance between the two frets is Δd = L₄ – L₃, so Δd = 1.09532 - 0.87033, leading to Δd = 0.22499 m or 22.5 cm, rounded to 23 cm.
6 0
2 months ago
On a foggy day, you are driving at 70 mph and Granny is driving at 50 mph. As you try to pass her, you both see an overturned tr
inna [3103]

Response:

C?

Reasoning:

7 0
3 months ago
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