Answer:
the object has a net force acting on it, it will accelerate the object will speed up, slow down, or change direction.
Explanation:
The reaction of radiodecay of carbon C-14 is
C-14 --> N-14 + e- + (ve)
where e- is an electron and (ve) is an electron-type antineutrino.
Basically, when the carbon nucleus (atomic number: 6, mass number: 14) decays, a neutron of the nucleus converts into a proton (therefore, the mass number remains the same, 14, but the atomic number increases by 1, therefore it becomes nitrogen) and releases an electron-antineutrino pair.
So, the correct answer is C), N-14.
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Answer:
Relation between initial speed of bullet and height h is given as
Explanation:
As we know that system of block and bullet swings up to height h after collision
So we have
so we have
so speed of the block + bullet just after the impact is given by above equation
Now we also know that there is no force on the system of bullet + block in the direction of motion
So we can use momentum conservation
now we have
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.
