The pitch of the sound of the car will appear to be <span>lower than the pitch of the car next to you</span>
ANSWER ALL QUESTIONS and show your work ON THE ATTACHMENT AWARD 50 pts if you don’t know the answer to all of them I have posted them individually so I can still mark you brainlessly (however you spell the darn thing)
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
Answer:
a).![v=83.77x10^{-3} rad/s](https://tex.z-dn.net/?f=v%3D83.77x10%5E%7B-3%7D%20rad%2Fs)
b).
rpm
c).v=0.5865 m/sec.
Explanation:
Given:
![r=8m](https://tex.z-dn.net/?f=r%3D8m)
![v=\frac{2rev}{2.5minutes}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2rev%7D%7B2.5minutes%7D)
a).
![2 rev*\frac{2\pirad}{1rev}=4\pi rad](https://tex.z-dn.net/?f=2%20rev%2A%5Cfrac%7B2%5Cpirad%7D%7B1rev%7D%3D4%5Cpi%20%20rad)
![t=2.5minutes*\frac{60s}{1minute}=150s](https://tex.z-dn.net/?f=t%3D2.5minutes%2A%5Cfrac%7B60s%7D%7B1minute%7D%3D150s)
The angular speed in radians per seconds is
![v=\frac{4\pi}{150s}=83.77x10^{-3} rad/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B4%5Cpi%7D%7B150s%7D%3D83.77x10%5E%7B-3%7D%20rad%2Fs)
b).
![v=\frac{2rev}{2.5minute}rpm](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2rev%7D%7B2.5minute%7Drpm)
rpm
c)
Child's distance per revolution
(pi*2r) = 43.988 metres.
v=(43.988 x 0.0133333) = 0.5865 m/sec.