Question

I’ll mark u BRAINLIEST,
IF U ANSWER THIS WITH EXPLANATION....

23rd AND 24th QUESTIONS PLEASE...

X NO SPAM X

Answer 1

$Answer \: \\ \\ for \: \: three \: concurrent \: coplanar \: forces \\ say \: \: f1 \: \: f2 \: \: and \: \: f3 \: \: we \: have \\ \\ \frac{f1}{ \sin( \alpha ) } = \frac{f2}{ \sin( \beta ) } = \frac{f3}{ \sin( \gamma ) } \\ where \: \: \alpha \: \: \: \beta \: \: \: and \: \gamma \: \: is \: the \: angle \: between \: \\( f2 \: f3) \: \: \: \: \: \: (f1 \: f3) \: \: and \: \: (f1 \: \: f2) \: \: respectively \: \\ \\ Question \: \: number \: \: \: 23 \\ \\ \frac{t1}{ \sin(90 + 60) } = \frac{t2}{ \sin(90) } = \frac{f}{ \sin(90 + 60) } \\ \\ \frac{t2}{ \sin(90) } = \frac{f}{ \sin(90 + 60) } \\ \\ \frac{t2}{ \sin(90) } = \frac{f}{ \cos(60) } \\ \\ t2 \: = \frac{f}{ \frac{ {1} }{2} } \\ \\ t2 = 2f \\ \\ therefore \: \: t2 = \: 2f\\ \\ Question \: number \: 24 \\ \\ \frac{t1}{ \sin(90 + 30) } = \frac{t2}{ \sin(90) } = \frac{ f }{ \sin(90 + 60) } \\ \\ \frac{t1}{ \cos(30) } = \frac{f}{ \cos(60) } \\ \\ \frac{t1}{ \frac{ \sqrt{3} }{2} } = \frac{f}{ \frac{1}{2} } \\ \\ \frac{2t1}{ \sqrt{3} } = 2f \\ \\ 2t1 = 2f \sqrt{3} \\ \\ t1 = f \sqrt{3} \\ \\ therefore \: \: t1 = f \sqrt{3} \\ \\ Note \: \\ \\ \sin(90 + x) = \cos(x)$

Answer 2

Answer:

According to your question n=3 which means third shell, i.e. 3s 3p 3d..

Now l=2 means d subshell. m=+2 comes only on d subshell..

s=+1/2 means 1 electron.

So finally the answer is 1