Explanation:
(a) Formula to calculate the density is as follows.
![\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7BQ%7D%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%20a%5E%7B3%7D%7D)
= ![\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}](https://tex.z-dn.net/?f=%5Cfrac%7B6.50%20%5Ctimes%2010%5E%7B-6%7D%7D%7B%5Cfrac%7B4%7D%7B3%7D%20%5Ctimes%203.14%20%5Ctimes%20%280.04%29%5E%7B3%7D%7D)
= ![2.42 \times 10^{-2} C/m^{3}](https://tex.z-dn.net/?f=2.42%20%5Ctimes%2010%5E%7B-2%7D%20C%2Fm%5E%7B3%7D)
Now, calculate the charge as follows.
![q_{in} = \rho(\frac{4}{3} \pi r^{3})](https://tex.z-dn.net/?f=q_%7Bin%7D%20%3D%20%5Crho%28%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%29)
= ![2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}](https://tex.z-dn.net/?f=2.42%20%5Ctimes%2010%5E%7B-2%7D%20C%2Fm%5E%7B3%7D%20%5Ctimes%204.1762%20%5Ctimes%20%280.01%29%5E%7B3%7D)
=
C
or, = 101.06 nC
(b) For r = 6.50 cm, the value of charge will be calculated as follows.
![q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}](https://tex.z-dn.net/?f=q_%7Bin%7D%20%3D%20%5Cfrac%7BQ%7D%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%20a%5E%7B3%7D%7D)
= ![\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}](https://tex.z-dn.net/?f=%5Cfrac%7B6.50%20%5Ctimes%2010%5E%7B-6%7D%7D%7B%5Cfrac%7B4%7D%7B3%7D%20%5Ctimes%203.14%20%5Ctimes%20%280.065%29%5E%7B3%7D%7D)
= 7.454 ![\mu C](https://tex.z-dn.net/?f=%5Cmu%20C)
Hello this is to other people looking for the answer. Everyone else is wrong. I just took the quiz on e2020. The answer actually Comparative. Your welcome. Have a nice day.
400m in 32sec: (400/32)>12.5meters per second>
(12.5)(60)(60)(1/1000)=45km per hour
Constant speed would mean that the two forces are equivalent
Answer:
1) 1.31 m/s2
2) 20.92 N
3) 8.53 m/s2
4) 1.76 m/s2
5) -8.53 m/s2
Explanation:
1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s
![a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D%20%3D%20%5Cfrac%7B17%20-%200%7D%7B13%7D%20%3D%201.31%20m%2Fs2)
2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration
![F_s = am = 1.31*16 = 20.92 N](https://tex.z-dn.net/?f=F_s%20%3D%20am%20%3D%201.31%2A16%20%3D%2020.92%20N)
3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.
![F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N](https://tex.z-dn.net/?f=F_%7B%5Cmu_s%7D%20%3D%20%5Cmu_sN%20%3D%20mg%5Cmu_s%20%3D%2016%2A9.81%2A0.87%20%3D%20136.6N)
So the maximum acceleration on the block is
![a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%20F_%7B%5Cmu_s%7D%20%2F%20m%20%3D%20136.6%20%2F%2016%20%3D%208.53%20m%2Fs%5E2)
4)As the box slides, it is now subjected to kinetic friction, which is
![F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N](https://tex.z-dn.net/?f=F_%7B%5Cmu_s%7D%20%3D%20mg%5Cmu_k%20%3D%2016%2A9.81%2A0.69%20%3D%20108.3%20N)
So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is
28.25 / 16 = 1.76 m/s2
5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2