Answer:
h f = Wf + K
where the total energy available is h f, Wf is the work function or the work needed to remove the electron and K is the kinetic energy of the removed electron
If K = zero then hf = Wf
Wf = h f = h c / λ or
λ = h c / Wf = 6.63E-34 * 3.0E8 / (3.7 * 1.6E-19)
λ = 6.63 * 3 / (3.7 * 1.6) E-7 = 3.36E-7
This would be 3360 angstroms or 336 millimicrons
Visible light = 400-700 millimicrons
Answer:
D
Explanation:
Let’s calculate the kinetic energy for all of the choices.
a. (1/2)(100)(100)^2 = 50(10000)=500,000
b. (1/2)(100)(1)^2 = 50
c. (1/2)(10)(100)^2 = 5(10000) = 50,000
d. (1/2)(1)(1)^2 = 0.5
We can see that (d) has the least kinetic energy.
Answer:
The total number of significant figures is twelve.
Explanation:
Answer:
About 66 miles per hour
Explanation:
Based on the information given we can assume the car traveled the same number of miles every hour meaning all we need to do is divide.
400/6 ≈ 66 miles per hour
Answer:
the distance from charge A to C is r₁₃= 1.216 m
Explanation:
following Coulomb's law , the force exerted by 2 point charges between themselves is:
F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant
since C ( denoted as 3) is at equilibrium
F₁₃=F₂₃
k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²
q₁/r₁₃²=q₂/r₂₃²
r₁₃²/q₁=r₂₃²/q₂
r₂₃=r₁₃*√(q₂/q₁)
since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have
r₁₃+r₂₃=d=r₁₂
r₁₃+r₁₃*√(q₂/q₁)=d
r₁₃*(1+√(q₂/q₁))=d
r₁₃=d/(1+√(q₂/q₁))
replacing values
r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m
thus the distance from charge A to C is r₁₃= 1.216 m
