Answer:So You Decide To Move Closer To Give The Conversation A Sound Level Of 80.0dB Instead. ... You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.0dB .
Explanation:
Answer:
1.6675×10^-16N
Explanation:
The force of gravity that the space shuttle experiences is expressed as;
g = GM/r²
G is the gravitational constant
M is the mass = 1.0 x 10^5 kg
r is the altitude = 200km = 200,000m
Substitute into the formula
g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²
g = 6.67×10^-6/4×10^10
g = 1.6675×10^{-6-10}
g = 1.6675×10^-16N
Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N
Answer:
The “terminal speed” of the ball bearing is 5.609 m/s
Explanation:
Radius of the steel ball R = 2.40 mm
Viscosity of honey η = 6.0 Pa/s
![\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Viscosity%20has%20Density%20%7D%20%5Csigma%3D1360%20%5Cmathrm%7Bkg%7D%20%2F%20%5Cmathrm%7Bm%7D%5E%7B3%7D)
![\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Steel%20has%20a%20density%20%7D%20%5Crho%3D7800%20%5Cmathrm%7Bkg%7D%20%2F%20%5Cmathrm%7Bm%7D%5E%7B3%7D)
![\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)](https://tex.z-dn.net/?f=%5Cleft.%5Cmathrm%7Bg%7D%3D9.8%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D%20%5Ctext%20%7B%20%28g%20is%20referred%20to%20as%20the%20acceleration%20of%20gravity.%20Its%20value%20is%20%7D%209.8%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D%5E%7B2%7D%20%5Ctext%20%7B%20on%20Earth%20%7D%5Cright%29)
While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)
![V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}](https://tex.z-dn.net/?f=V_%7Bt%7D%3D%5Cfrac%7B2%20%5Cmathrm%7BR%7D%5E%7B2%7D%28%5Crho-%5Csigma%29%20%5Cmathrm%7Bg%7D%7D%7B9%20%5Ceta%7D)
Substitute the given values to find "terminal speed"
![\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}](https://tex.z-dn.net/?f=%5Cmathrm%7BV%7D_%7B%5Cmathrm%7Bt%7D%7D%3D%5Cfrac%7B2%20%5Ctimes%200.0024%5E%7B2%7D%287800-1360%29%209.8%7D%7B9%20%5Ctimes%206%7D)
![\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}](https://tex.z-dn.net/?f=%5Cmathrm%7BV%7D_%7B%5Cmathrm%7Bt%7D%7D%3D%5Cfrac%7B0.0048%20%5Ctimes%206440%20%5Ctimes%209.8%7D%7B54%7D)
![\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}](https://tex.z-dn.net/?f=%5Cmathrm%7BV%7D_%7B%5Cmathrm%7Bt%7D%7D%3D%5Cfrac%7B302.9376%7D%7B54%7D)
![\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}](https://tex.z-dn.net/?f=%5Cmathrm%7BV%7D_%7B%5Cmathrm%7Bt%7D%7D%3D5.609%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D)
The “terminal speed” of the ball bearing is 5.609 m/s
We know that:
d=vt
d=32mph*5h
d=160mi
It will be traveling exactly 24 miles per hour <span />