Answer:
When the starting and ending points are the same, the total work is zero.
Explanation:
option ( D )correct
A force is said to be conservative when the work done by the force in moving a particle from a point A to a point B is independent of the path followed between A and B and is the same for all the paths. The work done depends only on the particles initial and final positions. And when the initial and final position in conservative field are same the work done is said to be zero.
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
The electrical potential energy of a charge q located at a point at potential V is given by

Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be

In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to

Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
Answer: there are 15 coins of $2 and 18 coins of $5
Explanation:
I will answer in English.
X is the number of $5 coins.
Y is the number of $2 coins.
We have the system of equations:
Y + X = 33
Y*2 + X*5 = 120
first, we must isolate one of the variables in one of the equations and then replace it in the other equation, let's isolate Y in the first equation:
Y = 33 - X.
Then we can replace it in the other equation:
(33 - X)*2 + X*5 = 120
66 - X*2 + X*5 = 120
X*3 = 54
X = 54/3 = 18
and using the equation for Y.
Y = 33 - X = 33 - 18 = 15
So there are 15 coins of $2 and 18 coins of $5
The half-life equation
in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>
<em>
</em>
<em>
</em>
Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.
<em>2 * 5700 = </em>11400 yr