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3 years ago

11

How many electrons do alkali metals have in their outer shell?

2 answers:

DIA [1.3K]3 years ago

6 0

These metals have only one electron<span> in their outer shell. Therefore, they are ready to lose that </span>one electron<span> in ionic bonding with other elements. #stay smoking

</span>

BARSIC [14]3 years ago

3 0

The outer shell can hold 1 electron

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The velocity of a car changes from 50 m/s north to 40 m/s north in 2 seconds. What is the car’s acceleration?

Zigmanuir [339]

Acceleration is defined as the change in velocity divided by the change in time. Since the velocity changed from 50 m/s to 40 m/s (-10 m/s change) in 2 seconds, then the acceleration is (-10/2) = -5 m/s^2 northward. Hope this helps!

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4 years ago

How do you calculate the refractive index of a material using the critical angle? (GCSE Physics)

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Answer:

µ = $\frac{1}{sinC}$

Explanation:

µ = 1/ sinC

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C ----> critical angle

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3 years ago

Calculate the change in internal energy (ΔE) for a system that is giving off 25.0 kJ of heat and is changing from 12.00 L to 6.0

Alexxandr [17]

Explanation:

The given data is as follows.

q = -25.0 kJ, Pressure (P) = 1.50 atm

$\Delta V$ = (12 - 6) L = 6 L

Therefore, product of pressure and change in volume will be as follows.

$P \Delta V = 1.50 atm \times 6 L$

= 7.5 L atm

= $7.5 \times 101.3$

= 759.75 J

Now, we will calculate the change in internal energy as follows.

$\Delta E = q + w$

= $q + P \Delta V$

= -25000 kJ + 759.75 J

= 24240.25 J

or, = 24.240 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the change in internal energy ( $\Delta E$ ) for a system is 24.240 kJ.

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4 years ago

In measuring the width of a hair sample, a light of wavelength 500 nm is used. The hair sample is 40 um in radius. With the scre

sergejj [24]

Answer:

The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

Explanation:

Given that,

Wave length = 500 nm

Radius $d= 40\ \mu m$

Distance from the hair sample D= 6 m

We need to calculate the distance of the second dark band away from the central bright spot be located

$\sin\theta=\dfrac{y}{D}$

$\sin\theta=\dfrac{y}{6}$

Using formula for dark fringe

$(n-\dfrac{1}{2})\lambda=2d\sin\theta$

Put the value into the formula

$(2-\dfrac{1}{2})\times500\times10^{-9}=2\times40\times10^{-6}\times\dfrac{y}{6}$

$y=\dfrac{(2-\dfrac{1}{2})\times500\times10^{-9}\times6}{2\times40\times10^{-6}}$

$y=0.05625\ m$

$y=5.625\times10^{-2}\ m$

Hence, The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

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B is the answerrrr so put it and get a A

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