Answer:
well, as u can tell the top layer will always be the youngest layer aka the newest layer. The farther u go down the older the layers get. So the deeper u dig the farther back in time we see.
Explanation:
Answer:
Shorter string produces more frequency in two different strings because the equation for frequency is velocity/wavelength , this means that a shorter string creates a shorter wavelength which essentially increases the total frequency produced
Answer:
The intensity at 10° from the center is 3.06 × 10⁻⁴I₀
Explanation:
The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ
I₀ = maximum intensity of light
a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m
θ = angle at intensity point = 10°
λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m
α = πasinθ/λ
= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m
= 1.0911/650 × 10³
= 0.001679 × 10³
= 1.679
Now, the intensity I is
I = I₀(sinα/α)²
= I₀(sin1.679/1.679)²
= I₀(0.0293/1.679)²
= 0.0175²I₀
= 0.0003063I₀
= 3.06 × 10⁻⁴I₀
So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀
Answer:
The horizontal range will be 
Explanation:
We have given initial speed of the shell u = 
Angle of projection = 51°
Acceleration due to gravity 
We have to find maximum range
Horizontal range in projectile motion is given by

So the horizontal range will be 