Answer is D - five.
<em>Explanation;</em>
- Electron dot diagrams show the valence electrons around the element by using dots.
- Valence electrons are the electrons which are in outermost shell of the atom.
-The atomic number of the N atom is 7.
Atomic number = number of protons = 7
If the atom is neutral,
number of protons = number of electrons.
Hence, N atom has 7 electrons.
- The electron configuration is 1s² 2s² 2p³.
Hence, N atom has 2 + 3 = 5 valence electrons. So, five electrons are represented in electron dot diagram of N.
Answer:
Explanation:
Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s
An impulse results in a change of momentum
The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s
The remaining momentum is 4.5e6 - 0.12e6 = 4.38e6 kg•m/s
The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s
The tug applies 0.012e6 N•s of impulse each second.
The initial barge momentum will be zero in
t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds
To stop the barge in one minute(60 s), the tug would have to apply
4.5e6 / 60 = 75000 N•s /s or 75 000 N
electric field lines are graphical presentation of electric field intensity
It is the graphical way to represent the electric field variation
If we draw the tangent to electric field line then it will give the direction of net electric field at that point
So whenever we draw the electric field lines of a charge distribution then it will always follow this basic properties
here we will always follow these basic properties of field lines
now as we can see that here two positive charges are placed nearby so the electric field must be like it can not intersect at any point because at intersection of two lines the direction of electric field not defined
As we have two directions of tangents at that point
So here the incorrect presentation is the intersection of two field lines which is not possible
Answer:
Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
![W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5E%7Bx_f%7D_%7Bx_0%7D%20%7BF%28x%29%7D%20%5C%2C%20dx%20%5C%5CW%3D%5Cint%5Climits%5E%7B14%7D_%7B0%7D%20%28%7B18N-0.530%5Cfrac%7BN%7D%7Bm%7Dx%7D%29%20%5C%2C%20dx%5C%5CW%3D%5B%2818N%29x-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7Bx%5E2%7D%7B2%7D%5D%5E%7B14%7D_%7B0%7D%5C%5CW%3D%2818N%2914m-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B%2814m%29%5E2%7D%7B2%7D-%2818N%290%2B%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B0%5E2%7D%7B2%7D%5C%5CW%3D252N%5Ccdot%20m-52N%5Ccdot%20m%5C%5CW%3D200N%5Ccdot%20m)
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:
The box is initially at rest, so 
. Solving for 
:
