Answer:
1.170*10^-3 m
3.23*10^-32 m
Explanation:
To solve this, we apply Heisenberg's uncertainty principle.
the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"
Δp * Δx = h/4π
m(e).Δv * Δx = h/4π
If we make Δx the subject of formula, by rearranging, we have
Δx = h / 4π * m(e).Δv
on substituting the values, we have
for the electron
Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 5.67*10^-31
Δx = 1.170*10^-3 m
for the bullet
Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 0.021
Δx = 3.23*10^-32 m
therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet
Answer:
r₁/r₂ = 1/2 = 0.5
Explanation:
The resistance of a wire is given by the following formula:
R = ρL/A
where,
R = Resistance of wire
ρ = resistivity of the material of wire
L = Length of wire
A = Cross-sectional area of wire = πr²
r = radius of wire
Therefore,
R = ρL/πr²
<u>FOR WIRE A</u>:
R₁ = ρ₁L₁/πr₁² -------- equation 1
<u>FOR WIRE B</u>:
R₂ = ρ₂L₂/πr₂² -------- equation 2
It is given that resistance of wire A is four times greater than the resistance of wire B.
R₁ = 4 R₂
using values from equation 1 and equation 2:
ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²
since, the material and length of both wires are same.
ρ₁ = ρ₂ = ρ
L₁ = L₂ = L
Therefore,
ρL/πr₁² = 4ρL/πr₂²
1/r₁² = 4/r₂²
r₁²/r₂² = 1/4
taking square root on both sides:
<u>r₁/r₂ = 1/2 = 0.5</u>
Answer:
load (l)=400N
Effort(E)=50N
mechanical advantage (MA)= load ÷Effort
(ma)=400÷50
(ma)=8
Explanation:
I copy pasted from the answer from the same question. Remember to first check if ur question is there
