Fc=mv^2/r so we get
2000kg*(25m/s)^2/(80m)= 15625N of force
hope this helps! Thank You!!
As the water plunges, its velocity increases. Its potential energy<span> becomes kinetic</span>energy<span>. The law of conservation of </span>energy<span> states that when one form of </span>energy<span> is</span>transformed<span> to another, no </span>energy<span> is destroyed in the process. ... So the total amount of </span>energy<span> is the same before and after any </span>transformation<span>.
hope it helps
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Answer:
Explanation:
Given
mass of archer 
Average force 
extension in arrow 
Work done to stretch the bow with arrow
This work done is converted into kinetic Energy of arrow
where v= velocity of arrow
(b)if arrow is thrown vertically upward then this energy is converted to Potential energy
No, options are given but I believe the answer would be
In a water cycle Solid state of matter has the particles closest together.
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.
