Answer: ![(48.41,\ 52.19)](https://tex.z-dn.net/?f=%2848.41%2C%5C%2052.19%29)
Explanation:
The confidence interval for population mean is given by :-
![\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%5Cpm%20z_%7B%5Calpha%2F2%7D%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
Given : Sample size : ![n=9](https://tex.z-dn.net/?f=n%3D9)
Sample mean : ![\ovreline{x}=50.3\text{ decibels}](https://tex.z-dn.net/?f=%5Covreline%7Bx%7D%3D50.3%5Ctext%7B%20decibels%7D)
Standard deviation : ![\sigma=2.9\text{ decibels }](https://tex.z-dn.net/?f=%5Csigma%3D2.9%5Ctext%7B%20decibels%20%7D)
Significance level : ![\alpha=1-0.95=0.05](https://tex.z-dn.net/?f=%5Calpha%3D1-0.95%3D0.05)
Critical value : ![z_{\alpha/2}=z_{0.025}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3Dz_%7B0.025%7D%3D1.96)
Now, the 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type :-
![50.3\pm (1.96)\dfrac{2.9}{\sqrt{9}}\\\\\approx50.3\pm1.89\\\\=(50.3-1.89,\ 50.3+1.89)=(48.41,\ 52.19)](https://tex.z-dn.net/?f=50.3%5Cpm%20%281.96%29%5Cdfrac%7B2.9%7D%7B%5Csqrt%7B9%7D%7D%5C%5C%5C%5C%5Capprox50.3%5Cpm1.89%5C%5C%5C%5C%3D%2850.3-1.89%2C%5C%2050.3%2B1.89%29%3D%2848.41%2C%5C%2052.19%29)
Answer:
![\omega = 0.016\,\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%200.016%5C%2C%5Cfrac%7Brad%7D%7Bs%7D)
Explanation:
The rotation rate of the man is:
![\omega = \frac{v}{R}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7Bv%7D%7BR%7D)
![\omega = \frac{0.80\,\frac{m}{s} }{5\,m}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B0.80%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%7D%7B5%5C%2Cm%7D)
![\omega = 0.16\,\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%200.16%5C%2C%5Cfrac%7Brad%7D%7Bs%7D)
The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:
![(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega](https://tex.z-dn.net/?f=%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%5Ccdot%20%280.16%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%29%20%3D%20%5B%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%2B20000%5C%2Ckg%5Ccdot%20m%5E%7B2%7D%5D%5Ccdot%20%5Comega)
The final angular speed is:
![\omega = 0.016\,\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%200.016%5C%2C%5Cfrac%7Brad%7D%7Bs%7D)
Answer:
3.71 m/s
Explanation:
From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as
since all the potential energy is transformed to kinetic energy
Making v the subject of the formula then
and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that
Answer:
using a formula
average velocity=total displacement/total time
average velocity=120/1minute40sec
1minute=60sec
60sec+40sec=120sec
average velocity=120m/120sec
average velocity=1m/sec