Answer:
width of slit =1.23× 10⁻⁶ m
Explanation:
we know the condition of diffraction minima,
d sin θ = n λ
λ = wavelength θ = angle between the central maxima and 1st minima
d = slit width
for first minima n = 1
now,
d =
d =
d = 1228 × 10⁻⁹ m = 1.228× 10⁻⁶ m
d = 1.23× 10⁻⁶ m
width of slit =1.23× 10⁻⁶ m
A i think because it’s talking about covering ground, a faster car would cover more ground.
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
Answer:
the phenomenon whereby a pair of particles are generated in such a way that the individual quantum states of each are indefinite until measured, and the act of measuring one determines the result of measuring the other, even when at a distance from each other.
Answer:
10.78 s
Explanation:
The force on the charge is computed by using the equation:
F = ma
∴
At time t(sec; the partiCle velocity becomes 
The velocity of the charge after the time t(sec) is expressed by using the formula:
