Explanation :
It is given that, parallel wave fronts are incident on an opening in a barrier. Diffraction of the waves depends on the size of opening and the wavelength of light used as per the following relation :
where,
is the wavelength of light
and a is the opening size (from fig)
So, it is clear that diffraction of waves depends directly on the wavelength of light and inversely on the size of opening.
Hence, the correct option is (2).
1) Example of contact force: friction
2) Examples of non-contact forces: gravity and electromagnetic force
Explanation:
1)
Contact forces are forces that acts only when the objects involved are touching.
An example of contact force in the geosphere is friction. Friction is a force that acts when two objects slide past each other, and the surfaces of the two objects are in contact. Due to the presence of "microbumps" on the two surfaces, there is a resistive force opposing the motion of the two objects, and this force is called friction.
Friction also acts when an object is moving through a fluid, although it takes a different name: resistance. Also in this case, the resistance acts in the direction opposite to the motion of the object, slowing it down.
2)
Non-contact forces are forces that act from a distance, therefore they act even when the objects involved are not touching.
Examples of non-contact forces are:
- Gravitational force: this is an attractive force that acts between any object with mass. Its magnitude is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between the two objects.
- Electromagnetic force: this is a force exerted between electrically charged objects. Its magnitude is directly proportional to the product of the two charges and inversely proportional to the square of the distance between the charges. It can be attractive (if the charges have opposite sign) or repulsive (if the charges have same sign).
Learn more about forces:
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Answer:
Explanation:
Given that:
If we let the piece of the close lose contact at ∠θ;
Then ; from force balance;
we have:
where;
Again:
Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.
Answer:
I = 8.75 kg m
Explanation:
This is a rotational movement exercise, let's start with kinetic energy
K = ½ I w²
They tell us that K = 330 J, let's find the angular velocity with kinematics
w² = w₀² + 2 α θ
as part of rest w₀ = 0
w = √ 2α θ
let's reduce the revolutions to the SI system
θ = 30.0 rev (2π rad / 1 rev) = 60π rad
let's calculate the angular velocity
w = √(2 0.200 60π)
w = 8.683 rad / s
we clear from the first equation
I = 2K / w²
let's calculate
I = 2 330 / 8,683²
I = 8.75 kg m
