The object has potential and kinetic energy when it jumps into the air
Answer:
It remains constant
Explanation:
As we know that buoyant force on an object given as
Fb = ρ Vd g
ρ= Density of fluid
Vd=Volume displace by body
g=10 m/s²
Fb =buoyant force
So from above we can say that buoyant force does not depends on the depth. It only depends on the fluid density and volume displace by body.
So when rock gets deeper and deeper the buoyant force will remain constant.
It remains constant
Answer:
V = V0 + a t
V = 75 - 9.8 * 3 = 45.6 m/s
Answer:
The earthquake occurred at a distance of 1122 km
Explanation:
Given;
speed of the P wave, v₁ = 8.5 km/s
speed of the S wave, v₂ = 5.5 km/s
The distance traveled by both waves is the same and it is given as;
Δx = v₁t₁ = v₂t₂
let the time taken by the wave with greater speed = t₁
then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.
v₁t₁ = v₂t₂
v₁t₁ = v₂(t₁ + 1.2 min)
v₁t₁ = v₂(t₁ + 72 s)
v₁t₁ = v₂t₁ + 72v₂
v₁t₁ - v₂t₁ = 72v₂
t₁(v₁ - v₂) = 72v₂

The distance traveled is given by;
Δx = v₁t₁
Δx = (8.5)(132)
Δx = 1122 km
Therefore, the earthquake occurred at a distance of 1122 km
But cracked cracks nuts while scripts cut