Answer:
a) 0.31 rad/s
b) 100 J
c) 6.67 W
Explanation:
(a) the force would generate a torque of:
![T = FR = 18 * 2.4 = 43.2 Nm](https://tex.z-dn.net/?f=T%20%3D%20FR%20%3D%2018%20%2A%202.4%20%3D%2043.2%20Nm)
According to Newton 2nd law, the angular acceleration would be
![\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7BT%7D%7BI%7D%20%3D%20%5Cfrac%7B43.2%7D%7B2100%7D%20%3D%200.021%20rad%2Fs%5E2)
It starts from rest, then after 15s it would achieve a speed of
![\omega = \alpha t = 0.021 * 15 = 0.31 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Calpha%20t%20%3D%200.021%20%2A%2015%20%3D%200.31%20rad%2Fs)
(b) The distance angle swept by it is:
![\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B%5Calpha%20t%5E2%7D%7B2%7D%20%3D%20%5Cfrac%7B0.021%20%2A%2015%5E2%7D%7B2%7D%20%3D%202.314%20rad)
Hence the work by the child
![W = T\theta = 43.2 *2.314 \approx 100 J](https://tex.z-dn.net/?f=W%20%3D%20T%5Ctheta%20%3D%2043.2%20%2A2.314%20%20%5Capprox%20100%20J)
c) Average power to work per time unit
![P = \frac{W}{t} = \frac{100}{15} = 6.67 W](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BW%7D%7Bt%7D%20%3D%20%5Cfrac%7B100%7D%7B15%7D%20%3D%206.67%20W)
Explanation:
It is given that,
The volume of a right circular cylindrical, ![V=108\ cm^3](https://tex.z-dn.net/?f=V%3D108%5C%20cm%5E3)
We know that the volume of the cylinder is given by :
![V=\pi r^2 h](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2%20h)
............(1)
The upper area is given by :
![A=32r^2+2\pi rh](https://tex.z-dn.net/?f=A%3D32r%5E2%2B2%5Cpi%20rh)
![A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}](https://tex.z-dn.net/?f=A%3D32r%5E2%2B2%5Cpi%20r%5Ctimes%20%5Cdfrac%7B108%7D%7B%5Cpi%20r%5E2%7D)
![A=32r^2+\dfrac{216}{r}](https://tex.z-dn.net/?f=A%3D32r%5E2%2B%5Cdfrac%7B216%7D%7Br%7D)
For maximum area, differentiate above equation wrt r such that, we get :
![\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}](https://tex.z-dn.net/?f=%5Cdfrac%7BdA%7D%7Bdr%7D%3D64r-%5Cdfrac%7B216%7D%7Br%5E2%7D)
![64r-\dfrac{216}{r^2}=0](https://tex.z-dn.net/?f=64r-%5Cdfrac%7B216%7D%7Br%5E2%7D%3D0)
![r^3=\dfrac{216}{64}](https://tex.z-dn.net/?f=r%5E3%3D%5Cdfrac%7B216%7D%7B64%7D)
r = 1.83 m
Dividing equation (1) with r such that,
![\dfrac{h}{r}=\dfrac{108}{\pi r}](https://tex.z-dn.net/?f=%5Cdfrac%7Bh%7D%7Br%7D%3D%5Cdfrac%7B108%7D%7B%5Cpi%20r%7D)
![\dfrac{h}{r}=\dfrac{108}{\pi 1.83}](https://tex.z-dn.net/?f=%5Cdfrac%7Bh%7D%7Br%7D%3D%5Cdfrac%7B108%7D%7B%5Cpi%201.83%7D)
![\dfrac{h}{r}=59 \pi](https://tex.z-dn.net/?f=%5Cdfrac%7Bh%7D%7Br%7D%3D59%20%5Cpi)
Hence, this is the required solution.
Answer:
The mass of Uranium present in a 1.2mg sample is ![4.8 \cdot 10^{-6}\,mg](https://tex.z-dn.net/?f=4.8%20%5Ccdot%2010%5E%7B-6%7D%5C%2Cmg)
Explanation:
The ration between Uranium mass and total sample mass is:
For a sample of mass 1.2 mg, the amount of uranium is:
![1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg](https://tex.z-dn.net/?f=1.2%5C%2C%20mg%20%5Ccdot%20%5Cfrac%7B1%7D%7B250000%7D%3D4.8%20%5Ccdot%2010%5E%7B-6%7D%5C%2Cmg)
Answer:
this description is valid for mediadle displacement, bone is an acceptable description
Explanation:
The description of a person's position must be done with a position vector. These vectors must have magnitude, a given direction and a starting point.
In the description this has a starting point corner NO of pine and 675.
Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.
After analyzing this description is valid for mediadle displacement, bone is an acceptable description
The mass of an atom comes from the protons and neutrons that is found in the nucleus. The number of protons is the atomic number of an element. To find the number of neutrons, subtract the atomic number from the mass of an atom. For example, sodium’s atomic number is 11. This will tell us that sodium has 11 protons in it. The atomic mass of sodium is 23. So subtract 23 form 11 gives us 12. Therefore, there are 12 neutrons in sodium.