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3 years ago

An electron is emitted from a nucleus- State the effect this has on the charge of the nucleus

Physics

2 answers:

abruzzese [7]3 years ago

8 0

The charge of the nucleus becomes positive.

The effect of an electron being removed has on the charge of the nucleus is actually quite simple.

Think about it like this. An atom has an equal amount of electrons and protons.

Say, Hydrogen.

Protons:1

Electrons:1

The charge is equal. Now, take away an electron.

Protons:1

Electrons:0

There is no negativity now, that an electron is absent.

This makes the charge of the nucleus <em>positive</em>, due to the absence of an electron.

Hope this helps and please consider marking branliest.

Likurg_2 [28]3 years ago

5 0

$Question$

An electron is emitted from a nucleus- State the effect this has on the charge of the nucleus

Answer:

an electron orbiting around the nucleus combines with a nuclear proton to produce a neutron, which remains in the nucleus, and a neutrino, which is emitted. As in positron emission, the nuclear positive charge and hence the atomic number decreases by one unit, and the mass number remains the same.

hope this helps!!!

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A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the

erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, $T_{b} = 75^{\circ}C$

Air temperature, $T_{infty} = 20^{\circ}$

k = 388 W/mK

h = $20\ W/m^{2}K$

Now,

Perimeter of the fin, p = $\pi D = 0.005\pi \ m$

Cross-sectional area of the fin, A = $\frac{\pi}{4}D^{2}$

A = $\frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}$

To calculate the heat transfer rate:

$Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})$

where

$m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237$

Now,

$Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W$

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3 years ago

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daser333 [38]

Answer: Gravity is the force that keeps planets in orbit around the Sun. Gravity alone holds us to Earth's surface.

Planets have measurable properties, such as size, mass, density, and composition. A planet's size and mass determines its gravitational pull.

A planet's mass and size determines how strong its gravitational pull is.

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Explanation:

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vagabundo [1.1K]

Answer:Expression given below

Explanation:

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$\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2$

$v=x\sqrt{\frac{k}{m_1}}$

now conserving momentum

$m_1v=\left ( m_1+m_2\right )v_0$

$v_0=\frac{m_1}{m_1+m_2}v$

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