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3 years ago

when an element tends to lose its valence electrons in chemical reactions , does it behave more like a metal or nonmetal

Physics

1 answer:

Juli2301 [7.4K]3 years ago

4 0

It behaves more like a metal

Explanation:

When an element tends to lose its valence electrons in chemical reactions, they behave more like a metal.

Metals are electropositive.

Electropositivity or metallicity is the a measure of the tendency of atoms of an element to lose electrons.

This is closely related to ionization energy and the electronegativity of the element.

The lower the ionization energy of an element, the more electropositive or metallic the element is .

Metals are usually large size and prefers to be in reactions where they can easily lose their valence electrons.

When most metals lose their valence electrons, they attain stability.

Non-metals are electronegative. They prefer to gain electrons.

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What is the weight of a basketball with a mass of 0.5 kg

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3 years ago

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

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Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

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$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

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=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$

=> $\rho_d = 50.6*10^{-12} * 1.5*10^{3}$

=> $\rho_d = 7.59*10^{-8} \ C/m$

Generally the surface charge density is mathematically represented as

$\rho_s = \frac{\rho_l }{2 \pi * r_2 }$ here $2 \pi r = (circumference \ of \ outer \ conductor )$

=> $\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }$

=> $\rho_s = 4.026 *10^{-6} \ C/m^2$

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2 years ago

I need help with this question I’m not sure where to even begin

Strike441 [17]

We will have the following:

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We know that the system must be at equilibrium on the horizontal axis:

$\sum F_x=T_1cos(42.5)+T_2cos(36.5)=0$

So:

$\begin{gathered} T_2cos(36.5)=-(1235N)cos(42.5)\Rightarrow T_2=-\frac{(1235N)cos(42.5)}{cos(36.5)} \\ \\ \Rightarrow T_2=1132.711003...N\Rightarrow T_2\approx1132.7N \end{gathered}$

So, the value of T2 is approximately 1132.7 N.

b. We will determine the torques created by T1 and T2 as follows:

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$\tau_{T1}=(10m)(1235N)sin(42.5)\Rightarrow\tau_{T1}\approx8343.5N\ast m$

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$\tau_{T2}=(10m)(1132.7N)sin(36.5)\Rightarrow\tau_{T2}\approx6737.6N\ast m$

So the torques of T1 and T2 on the base are approximately 8343.5 N*m and 6737.6 N*m respectively.

c. The torques around that axis generated by the normal force and the weight are both 0 N*, since they are parallel to the axis.

d. We will determine the angular acceleration as follows:

$\begin{gathered} \alpha=\frac{\tau_{T1}}{I}\Rightarrow\alpha=\frac{\tau_{T1}}{(1/3mL^2)} \\ \\ \Rightarrow\alpha=\frac{(8343.5N\ast m)}{(1/3(200kg)(10m)^2)}\Rightarrow\alpha=1.251525rad/s^2 \\ \\ \Rightarrow\alpha\approx1.25rad/s^2 \end{gathered}$

So, the angular acceleration is approximately 1.25 radians/ s^2.

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