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:
Explanation:
From the conservation of mechanical energy
Solve to velocity v2
The angular momentum of an object is equal to the product of its moment of inertia and angular velocity.
L = Iω
I = 1/2 MR²
I = 1/2 x 13 x (0.2)
I = 1.3
ω = 2π/t
ω = 2π/0.3
ω = 20.9
L = 1.3 x 20.9
= 27.2 kgm²/s
As we know by the first law of thermodynamics
here we know that
Q = heat given to the system
W = work done by the system
now here we can say
now we can say that heat will be given as
now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy
So best describe Jin's Error is
<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>
Answer:
It comes out the positive side of the battery and goes in to the negative side of the battery
Explanation:
There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.
In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.
However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative
