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Describe the importance of the neutron in a atomic nuclei

RUDIKE [14]

Neutrons are required for the stability of nuclei, with the exception of the single-proton hydrogen nucleus. Neutrons are produced copiously in nuclear fission and fusion. They are a primary contributor to the nucleosynthesis of chemical elements within stars through fission, fusion, and neutron capture processes.

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2 years ago

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Four charges of equal magnitude q = 2.16 µC are situated as shown in the diagram below. If d = 0.88 m, find the electric potenti

Triss [41]

Answer:

The total potential (magnitude only) is 11045.45 V

Explanation:

Electric Potential

The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

$\displaystyle V=\frac{kq}{r}$

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

$\displaystyle V_1=\frac{9\cdot 10^{9}\times -2.16\cdot 10^{-6}}{0.88}=-22090.91\ V$

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

$V_2=-22090.91\ V$

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

$V_3=+22090.91\ V$

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

$\displaystyle V_4=\frac{9\cdot 10^{9}\times 2.16\cdot 10^{-6}}{1.76}=+11045.45 \ V$

The total electric potential in A is

$V=-22090.91\ V-22090.91\ V+22090.91\ V+11045.45 \ V$

$V=-11045.45 \ V$

The total potential (magnitude only) is 11045.45 V

8 0

3 years ago

The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig

Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

the upper and the lower channels are at the same pressure (the atmospheric pressure).
the velocity of water in the upper channel is zero (v₁ = 0 m/s).
y₁ = 2.00 m (height of the upper channel)
y₂ = 0.00 m (height of the lower channel)
g = 9.81 m/s²
ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A ⇒ A = Q/v ⇒ A = Q/v₂

⇒ A = (350 m³/s)/(6.264 m/s)

⇒ A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4 ⇒ D = 2*√(A/π)

⇒ D = 2*√(55.873 m²/π)