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3 years ago

5

How is electron movement related to the bonding in sodium chloride?

2 answers:

irina1246 [14]3 years ago

8 0

I think its c sorry if I’m wrong

Svetllana [295]3 years ago

7 0

I am pretty sure it is C, the other ones don’t make sense.

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It is number 3 because I know it is

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An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

So, 14 ms is required to reach the potential of 1500 V.

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3 years ago

Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210

kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

Given

Wavelength of light $(\lambda )=565nm \approx 565\times 10^{-9}m$

Width of slit(b)=0.210

(a)Width of central maximum located 1.80m from slit

$=\frac{2\lambda L}{b}$

$=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

=9.685 mm

(b)Width of the first order bright fringe

$Y_1=\frac{\lambda \times L}{b}$

$Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

$Y_1=4.84mm$

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Visible light energy.

Ipatiy [6.2K]

What about it??

Please explain and I will help.

I don’t know what you’re asking.

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