Answer:
electrons can be knocked loose from one object and picked up by the other. The object that gains electrons becomes negatively charged, while the object that loses electrons becomes positively charged.
Answer:
Electric Field = 3.369 x 10^4 N/C
Explanation:
Radius = r = (r1 + r2) / 2 = (1.6 + 3.6) /2 = 2.6 cm + 2.3 cm = 4.9 cm = 0.049 m
As we know, Electric field = E = kQ/r.r
= 8.98755 x 10^9 x 9 x 10^-9 / 0.049 x 0.049 = 33689.275 N/C
= 3.369 x 10^4 N/C
(c) When the two pulses completely overlap on the string forms a straight line.
A single disturbance that travels via a transmission medium is referred to as a pulse. This medium might be formed of stuff or a vacuum, and it might be endlessly large or finite in size.
Consider two pulses that are identical in shape and proceed in opposite directions along a string, with the exception that one has positive displacements of the string's elements while the other has negative displacements.
On the string, the two pulses blend together completely.
The pulses completely balance one another out in terms of removing string elements from equilibrium, yet the string still moves. Shortly after the string is once again shifted, the pulses will have passed each other.
The correct option is (c)
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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.
