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Alenkinab [10]
3 years ago
10

Which is not part of nuclear chemistry? A. Strong nuclear force B. Double-replacement reactions C. Radioactivity D. Nuclear deca

y
Physics
2 answers:
ziro4ka [17]3 years ago
8 0

The correct choice is

B. Double-replacement reactions

in this type of reaction , two ionic compounds react and swaps their cations or anions giving two new compounds in the product. hence the double replacement reactions does not deal with the nucleus. all other terms like strong nuclear forces, radioactivity and nuclear decay deal with the nucleus.

andrezito [222]3 years ago
7 0
D. Hope this helpsssss
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Kara is writing a science-fiction story. In the story, Earth no longer experiences day and night.
Ludmilka [50]

Answer:

C. She should get the Earth spinning on its axis again.

Explanation:

The Earth experiences day and night because of its spinning on its own axis.

If Kara's science-fiction story doesn't have the Earth spinning on its own axis, then the Earth will not experience day and night and hence Kara should incorporate this idea into his story.

6 0
3 years ago
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Can you tell from the coefficient of restitution whether a collision has added kinetic energy to a system, taken some away, or l
Maurinko [17]
For an inelastic collision where coefficient of restitution,e, is equal to 0, the momentum is conserved but not the kinetic energy. So, there is addition or elimination of kinetic energy.

On the otherhand, when e = 1, like for an elastic collision, kinetic energy and momentum is conserved. Thus, the system's kinetic energy is unchanged.
6 0
3 years ago
A long, solid rod 5.3 cm in radius carries a uniform volume charge density. Part A If the electric field strength at the surface
Blizzard [7]

Answer:

\rho = 7.35\times \muC/m^3

Explanation:

given,

Radius of the solid rod, R = 5.3 cm

Electric field strength,E = 22 kN/C

Let the volume charge density be ρ

From Gauss law

  E = \dfrac{Rl}{2\epsilon_0}

 ε₀ is the permitivity of free space

  R is the radius of the rod

 and also,

\rho=\dfrac{2E\epsilon_0}{R}

ρ is the volume charge density

\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}

\rho = 7.35\times 10^{-6}\ C/m^3

\rho = 7.35\times \muC/m^3

Hence, the volume charge density is equal to \rho = 7.35\times \muC/m^3

6 0
3 years ago
Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

  • M the mass of the planet, and
  • G the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0) <em>(equation 3)</em>

<em>Equation 4 </em> gives the value of gravitational acceleration, g, a point of negligible mass experiences at a distance r from a planet of mass M (assuming no other stellar object were present)

g = G \cdot M/ r^{2} <em>(equation 4)</em>

where the universal gravitation <em>constant</em> G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}

Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

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3 years ago
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