A surface wave is a wave that travels along the surface of a medium. The medium is the matter through which the wave travels. Ocean waves are the best-known examples of surface waves. They travel on the surface of the water between the ocean and the air. (According to ck12.org)
So your answer would be A!
Answer:
Triples
Explanation:
Kinetic energy =1/2 mv^2 + 1/2 Iw^2
v= linear velocity
w= angular velocity
I= moment of inertia of object
I = k mr^2 where k is a constant for a particular shaped object
v = rw
w=v/r
Rotational Kinetic energy = 1/2 k m× (v/r)^2 × r^2
Total Kinetic energy = translational Kinetic energy + rotational Kinetic energy
KE = 1/2 (1+k) mv^2
Centripetal acceleration = mv^2/r
When KE is tripled, because 1/2(1+k) is a constant for a particular object, mv^2 part gets tripled. r doesn't appear in the equation of total Kinetic energy. So tripling Kinetic energy doesn't affect r. Therefore centripetal acceleration also gets tripled.
Explanation:
I don't know what values you have measured, or what <em>Ladybug Revolution</em> is, but I can show you the equations for rotational motion.
Angular displacement is:
Δθ = ω₀ t + ½ αt²
where ω₀ is the initial angular velocity, α is the angular acceleration, and t is time.
Assuming the radius is constant, the arc length is radius times angular displacement.
s = rΔθ
Using the equation for angular displacement, we can see that if two points have the same angular velocity and same angular acceleration, then they also have the same angular displacement.
Answer:
θ_c = 24.4º
Explanation:
To find the critical angle, let's use the law of refraction where index 1 refers to the incident medium (diamond) and index 2 refers to the medium where it is to be refracted (air)
n₁ sin θ₁ = n₂ sin θ₂
for the critical angle the ray comes out refracted parallel to the surface, therefore the angle is
θ₂ = 90
n₁ sin θ₁ = n₂
θ_c = θ₁ = sin⁻¹ 
the index of refraction of the diamond is tabulated
n₁ = 2.419
let's calculate
θ_c = sin⁻¹ (
)
θ_c = 24.4º
Answer and Explanation:
As per the question:
When the stone is thrown from the cliff top and hits the ground below eventually:
R = 
where
= initial velocity
H = height
g = acceleration due to gravity
R = horizontal Range
Now,
(a) Displacement of the stone is given by the horizontal range:
R =
where
= initial velocity
H = height
g = acceleration due to gravity
R = horizontal Range
(b) Speed just prior to the impact is given by the third equation of motion:
where
v = final velocity
(c) Time of flight is given by the second eqn of motion where the initial velocity is considered to be 0 then:
T = 
