<h3><u>Minimum uncertainty in the vertical component of the momentum of each photon:</u></h3>
According to Heisenberg's Uncertainty principle, both the “position and velocity of the particle” cannot be measured exactly at the same time. The momentum of the particle equals the product of its mass and velocity. And it can be inferred that the “product of the uncertainties” in the “momentum and the position” of a particle equals 
.
Immediately after the photon has passed through the slit, given particle has a momentum uncertainty of 
and its position uncertainty is 
, then the minimum uncertainty in its momentum will be
Answer:
Given:
Object distance (u) = 25.0 cm
Image distance (v) = -50.0 cm
To Find:
Magnification (m)
Explanation:
Substituting values of Image distance(v) & Object distance (u) in the equation:
-(-50) = 50:
Answer:
716.1 m
Explanation:
A car was travelling at a velocity of 8m/s
It accelerated at 5m/s^2
It finally reached a velocity of 85m/s
The distance can be calculated by applying the fourth equation of motion
V^2= U^2 +2as
V= 85m/s
U= 8m/s
a= 5m/s^2
85^2= 8^2 + 2(5)(s)
7225= 64 + 10s
7225-64= 10s
7161=10s
s= 7161/10
= 716.1 m
Hence the car travelled at a distance of 716.1 m
