Explanation:
From the experiment:
Number of protons = 74
Number of neutrons = 110
Number of protons in an element is the atomic number of the element. It is used to locate and position and element on the periodic table.
For a neutral or uncharged atom, the number of protons is the same as the number of electrons.
The element whose number of protons or atomic number if 74 is Tungsten
Mass number = 74 + 110 = 184g/mol
Answer:
B
Explanation:
Transformation of energy involves conversion of energy from one form to another for example our movement around involves the conversion of chemical energy stored in the food we eat to other forms of energy such as kinetic energy for the movement, electrical energy in the neurons for impulses and others
The ball posses gravitational potential energy since it is held at a displacement to the ground ( zero point) and when released, the gravitational potential energy is converted to kinetic energy which leads to the fall of the ball until it is at zero displacement to the earth. The board likewise when bent to its maximum extent stored elastic potential energy as a result of the partial displacement of its constituent particle provided it is not stretch beyond its elastic limit which can lead to deformation of the board and the elastic potential energy lost.
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]