If the length of the wire increases, then the amount of resistance will also increase.
1. Take a long piece of wire and cut it 10 pieces. Those pieces should all be different sizes, one should be 5___ (units in meter, cm, inches, etc.), and the next should be 5 ___ (units in meter, cm, inches, etc.) more than the one before.
2. Take one piece of wire and measure the resistance using ___ and record the results in the data table.
3. Repeat the previous step with all the pieces of wire.
4. Compare and contrast the results you have found.
I hope this helps a bit :)
Answer:
explained
Explanation:
When the intensity of light is increased on a piece of metal only the number of electron ejected will increase because all other things independent of intensity of light.
Light below certain frequency will not cause any electron emission no matter how intense.
The intensity produces more electron but does not change the maximum kinetic energy of electrons.
Work function is independent of the intensity of light, because it is an intrinsic property of a material.
Answer:
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Explanation:
Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.
Substituting the values of the variables into the equation, we have
ΔV = V₂ - V₁.
ΔV = 175 V - 33 V.
ΔV = 142 V
The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.
So, substituting the values of the variables into the equation, we have
ΔU = eΔV
ΔU = eΔV
ΔU = -1.602 × 10⁻¹⁹ C × 142 V
ΔU = -227.484 × 10⁻¹⁹ J
ΔU = -2.27484 × 10⁻²¹ J
ΔU ≅ -2.275 × 10⁻²¹ J
So, the required equation for the electric potential energy change is
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Answer:
Rutherford and atomic model are correctly matched.
