Answer:
Protium
Explanation: Protium is an isotope of hydrogen that is composed of one proton and one electron. It is the most abundant form of hydrogen.
<span>We first calculate the velocity of the ball when it hits the ground; this is equal to the square root of the quantity (2*g*d) where g is the acceleration of gravity (9.8 m/s^2) and d is the distance fallen, 1.5m.
So, we get a velocity of sqrt(2*9.8*1.5) = 5.42 m/s.
We can calculate the impulse force applied to the putty ball by using Newton's second law, which states that the applied force is equal to the product of mass and acceleration, where acceleration can be further decomposed as the change in velocity divided by the change in time. Thus, inputting the known values, we have:
F = ma = m(dv/dt) = 1.0*5.42/0.045 = 120.4 newtons.</span>
Explanation:
It is given that,
Length of rod, l = 14 cm = 0.14 m
Total charge, 
We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m
Electric field at the axis of rod is given by :
Where
is the linear charge density, 
So, 
E = −7849988.22 N/C
or
Negative sign shows the direction of electric field is inward in all direction. Hence, this is the required solution.
The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is
<span>dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) </span>
<span>where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. </span>
<span>However, cos(T) = x/sqrt(x^2 + R^2) </span>
<span>so the equation becomes </span>
<span>dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) </span>
<span>dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 </span>
<span>Integrating around the ring you get </span>
<span>E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 </span>
<span>E = (R/2*e0)*x*(x^2 + R^2)^-1.5 </span>
<span>we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} </span>
<span>to find the maxima set this = 0, giving </span>
<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>
<span>mult both side by (x^2 + R^2)^2.5 to get </span>
<span>(x^2 + R^2) - 3*x^2 = 0 </span>
<span>-2*x^2 + R^2 = 0 </span>
<span>-2*x^2 = -R^2 </span>
<span>x = (+/-)R/sqrt(2) </span>
