<span>Rayed craters-</span><span> were the last features to form on the moon.</span>
The efficiency of the scissor is 200%.
<u>Explanation:</u>
Efficiency is defined as the ratio of output of any instrument or device or machine to the input supplied to it. So the greater the output the greater will be the efficiency of the device.
As here the work done by us on the system is said to be 10 J so this will be equal to the input work done on the system. And the work done by the system i.e., the scissor is 20 J, so this will be the output work.
So, the efficiency is the ratio of output to input as shown below.
Efficiency =
= 200
So, the efficiency of the scissor is 200%.
Answer
Explanation:
The question was incomplete as the events are not given in the question. However the answer to your question is given as follows. The correct order of the events from youngest (top) to oldest (bottom) is given as follows.
Moon formation
↑
Earth formation
↑
Nuclear fusion in protosun
↑
BigBang
Answer:
x=0.53![x10^{-3} m](https://tex.z-dn.net/?f=x10%5E%7B-3%7D%20m)
Explanation:
Using Gauss law the field is uniform so
E=ζ/ε
Charge densities ⇒ζ=1.![x10x^{-6} \frac{C}{m^{2}}](https://tex.z-dn.net/?f=x10x%5E%7B-6%7D%20%5Cfrac%7BC%7D%7Bm%5E%7B2%7D%7D)
ε=8.85![x10^{-12} \frac{C^{2}}{n*m^{2}}](https://tex.z-dn.net/?f=x10%5E%7B-12%7D%20%5Cfrac%7BC%5E%7B2%7D%7D%7Bn%2Am%5E%7B2%7D%7D)
![E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1x10%5E%7B-6%7D%5Cfrac%7BC%7D%7Bm%5E%7B2%7D%7D%7D%7B8.85x%5E%7B-12%7D%5Cfrac%7BC%5E%7B2%7D%20%7D%7BN%2Am%5E%7B2%7D%7D%7D%20%5C%5CE%3D0.11299%20x10%5E%7B-6%7D%20%5Cfrac%7BN%7D%7BC%7D)
Force of charge is
![F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N](https://tex.z-dn.net/?f=F_%7Bq%7D%3Dq%2AE%5C%5CF_%7Bq%7D%3D1.6x10%5E%7B-19%7DC%2A0.11299x10%5E%7B6%7D%5Cfrac%7BN%7D%7BC%7D%20%5C%5CF_%7Bq%7D%3D1.807x10%5E%7B-14%7D%20N)
![F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s](https://tex.z-dn.net/?f=F_%7Bq%7D%3Dm%2Aa%5C%5Ca%3D%5Cfrac%7BF_%7Bq%7D%7D%7Bm%7D%3D%5Cfrac%7B1.807x10%5E%7B-13%7DN%7D%7B1.67x10%5E%7B-27%7D%7D%5C%5C%20a%3D1.082x0%5E%7B14%7D%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%5C%5C%20x%3D2.1cm%5Cfrac%7B1m%7D%7B100cm%7D%3D0.021m%20%5C%5Cv%3D6.7x10%5E%7B6%7D%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5C%20t%3D%5Cfrac%7B0.021m%7D%7B6.7x10%5E%7B6%7D%5Cfrac%7Bm%7D%7Bs%7D%7D%20%5C%5Ct%3D3.13x10%5E%7B-9%7Ds)
So finally knowing the acceleration and the time the distance can be find using equation of uniform motion
![x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2} \\x_{f}=0.53x^{-3}m](https://tex.z-dn.net/?f=x_%7Bf%7D%3Dx_%7Bo%7D%2B%5Cfrac%7B1%7D%7B2%7D%2Aa%2At%5E%7B2%7D%5C%5C%20x_%7Bo%7D%3D0%5C%5Cx_%7Bf%7D%3D%5Cfrac%7B1%7D%7B2%7D%20a%2At%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7D%2A1.082x10%5E%7B14%7D%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%2A%283.134x%5E%7B-9%7Ds%29%5E%7B2%7D%20%20%5C%5Cx_%7Bf%7D%3D0.53x%5E%7B-3%7Dm)
Answer:
I believe the answer would be C. point z