Using
F= mv²/r
4 = 0.5×v² / 2
8 /0.5 = v²
v²=16
v= √16
v= 4 ms-¹
Answer:
0.02 s
Explanation:
Take the (+x) direction to be up.
The average velocity v during a time interval Δt is the displacement Δx divided by Δt.
v=Δx/Δt
=x_f-x_i/t_f-t_i (1)
We assume that your height is 1.6m
Solving [1]
Δt=Δx/v
= 0.02 s
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:
A
Explanation:
because u are subtracting if this is from flvs that is what i did and it was right
Answer:
The higher the amplitude, the higher the energy. To summarise, waves carry energy. The amount of energy they carry is related to their frequency and their amplitude. The higher the frequency, the more energy, and the higher the amplitude, the more energy.
