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
Answer:
I'm not sure if you maybe forgot to add in some information but I can give you the equation you will need to solve this..! (just without the bit of infomration you might need)
In this case you'll be solving for t and I can help you with your first step, you'll divide 2 by 4, getting you to:
Answer:
A. 98,000 J
Explanation:
The gravitational potential energy of an object is given by
U = mgh
where
m is the mass of the object
g is the gravitational acceleration
h is the heigth above the ground
In this problem,
m = 2000 kg
g = 9.8 m/s^2
h = 5.0 m
Substituting into the equation, we find