Density<span> is the </span>mass<span> of an object </span>divided<span> by its </span>volume<span>. So the answer would be Yes. Hope it helps! (:</span>
The net charge on an atom is equal to the overall difference between the number of protons in the nucleus versus the number of electrons around the nucleus, where a negative sign represents less protons and a positive sign represents more protons (than electrons).
The answer is "False". The force acting on the object is 27 N.
According to Newton's second law, when a force <em>F</em> acts on am object of mass <em>m</em>, it produces an acceleration <em>a</em>. The force is given by the expression,
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,
![F=ma\\ =(9.0kg)(3m/s^2)\\ =27N](https://tex.z-dn.net/?f=F%3Dma%5C%5C%20%3D%289.0kg%29%283m%2Fs%5E2%29%5C%5C%20%3D27N)
Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.
Answer:
Wavelength, ![\lambda=1.28\times 10^{-14}\ m](https://tex.z-dn.net/?f=%5Clambda%3D1.28%5Ctimes%2010%5E%7B-14%7D%5C%20m)
Explanation:
Given that,
Mass of the particle, ![m=4.3\times 10^{-28}\ kg](https://tex.z-dn.net/?f=m%3D4.3%5Ctimes%2010%5E%7B-28%7D%5C%20kg)
Acceleration of the particle, ![a=2.4\times 10^7\ m/s^2](https://tex.z-dn.net/?f=a%3D2.4%5Ctimes%2010%5E7%5C%20m%2Fs%5E2)
Time, t = 5 s
It starts from rest, u = 0
The De Broglie wavelength is given by :
![\lambda=\dfrac{h}{mv}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bh%7D%7Bmv%7D)
v = a × t
![\lambda=\dfrac{h}{mat}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bh%7D%7Bmat%7D)
![\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7B6.67%5Ctimes%2010%5E%7B-34%7D%7D%7B4.3%5Ctimes%2010%5E%7B-28%7D%5Ctimes%202.4%5Ctimes%2010%5E7%5Ctimes%205%7D)
![\lambda=1.28\times 10^{-14}\ m](https://tex.z-dn.net/?f=%5Clambda%3D1.28%5Ctimes%2010%5E%7B-14%7D%5C%20m)
Hence, this is the required solution.
D. used by the entire scientific community
B. more accurate system of measurement