Answer:
![v=12.65\ m.s^{-1}](https://tex.z-dn.net/?f=v%3D12.65%5C%20m.s%5E%7B-1%7D)
Explanation:
Given:
- mass of vehicle,
![m=1350\ kg](https://tex.z-dn.net/?f=m%3D1350%5C%20kg)
- radius of curvature,
![r=71\ m](https://tex.z-dn.net/?f=r%3D71%5C%20m)
- coefficient of friction,
![\mu=0.23](https://tex.z-dn.net/?f=%5Cmu%3D0.23)
<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>
![m.\frac{v^2}{r} =\mu.N](https://tex.z-dn.net/?f=m.%5Cfrac%7Bv%5E2%7D%7Br%7D%20%3D%5Cmu.N)
where:
coefficient of friction
normal reaction force due to weight of the car
velocity of the car
![1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)](https://tex.z-dn.net/?f=1350%5Ctimes%20%5Cfrac%7Bv%5E2%7D%7B71%7D%20%3D0.23%5Ctimes%20%281350%5Ctimes%209.8%29)
is the maximum velocity at which the vehicle can turn without skidding.
I believe the correct answer would be kimberlite. Diamonds are usually found in pipes 50 to 200 m across made of kimberlite. It is an igneous rock that is known to contain traces of diamonds. It is named base on the town where it was discovered which is Kimberley, South Africa.
From that list, only the frequency makes the difference.
Einstein won his only Nobel Prize for his explanation of this effect.
The force needed to accelerate a vehicle with a mass of 1000kg at a rate of 5m/s2 would be 5000
Answer:
Part a)
![a_c = 1.67 \times 10^{-10} m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%201.67%20%5Ctimes%2010%5E%7B-10%7D%20m%2Fs%5E2)
Part b)
![v = 2.18 \times 10^5 m/s](https://tex.z-dn.net/?f=v%20%3D%202.18%20%5Ctimes%2010%5E5%20m%2Fs)
Explanation:
Time period of sun is given as
![T = 2.60 \times 10^8 years](https://tex.z-dn.net/?f=T%20%3D%202.60%20%5Ctimes%2010%5E8%20years)
![T = 2.60 \times 10^8 (365 \times 24 \times 3600) s](https://tex.z-dn.net/?f=T%20%3D%202.60%20%5Ctimes%2010%5E8%20%28365%20%5Ctimes%2024%20%5Ctimes%203600%29%20s)
![T = 8.2 \times 10^{15} s](https://tex.z-dn.net/?f=T%20%3D%208.2%20%5Ctimes%2010%5E%7B15%7D%20s)
Now the radius of the orbit of sun is given as
![R = 3.00 \times 10^4 Ly](https://tex.z-dn.net/?f=R%20%3D%203.00%20%5Ctimes%2010%5E4%20Ly)
![R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m](https://tex.z-dn.net/?f=R%20%3D%203.00%20%5Ctimes%2010%5E4%20%283%5Ctimes%2010%5E8%29%28365%20%5Ctimes%2024%20%5Ctimes%203600%29m)
![R = 2.84 \times 10^20 m](https://tex.z-dn.net/?f=R%20%3D%202.84%20%5Ctimes%2010%5E20%20m)
Part a)
centripetal acceleration is given as
![a_c = \omega^2 R](https://tex.z-dn.net/?f=a_c%20%3D%20%5Comega%5E2%20R)
![a_c = \frac{4\pi^2}{T^2} R](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cfrac%7B4%5Cpi%5E2%7D%7BT%5E2%7D%20R)
![a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cfrac%7B4%5Cpi%5E2%7D%7B%288.2%5Ctimes%2010%5E%7B15%7D%29%5E2%7D%282.84%20%5Ctimes%2010%5E%7B20%7D%29)
![a_c = 1.67 \times 10^{-10} m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%201.67%20%5Ctimes%2010%5E%7B-10%7D%20m%2Fs%5E2)
Part b)
orbital speed is given as
![v = \frac{2\pi R}{T}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%20R%7D%7BT%7D)
![v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%20%282.84%20%5Ctimes%2010%5E%7B20%7D%29%7D%7B8.2%20%5Ctimes%2010%5E%7B15%7D%7D)
![v = 2.18 \times 10^5 m/s](https://tex.z-dn.net/?f=v%20%3D%202.18%20%5Ctimes%2010%5E5%20m%2Fs)