Question

✨Physics Genius✨
>Class 11
>Solve mainly for part b) & c)

No Spamming.
Answer Genuinely.
​

Answer 1

Hey guy here your answer....

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Answer 2

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge\mathcal{\bf{{\underline{\underline{\huge\mathcal{...Answer...}}}}}}$

$\underline{\large\mathcal\red{solution}}$

‼️✓✓Theorem

Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector

‼️✓✓Formula

AB+BC=AC. Or. P+Q=R

‼️✓✓Proof:

Now ,a perpendicular is drawn on the extended portion of the line AB From the point C.

i.e., <CDA=90°

Consider two vectors P and Q that represent the magnitude and direction of the sides AB and BC respectively of the triangle ABC. Let R be the resultant of vectors P and Q.

Now...,

$</p><p>(AC){}^{2}=(AD){}^{2}+(CD){}^{2}\\</p><p>=>(AC){}^{2}=(AB+BD){}^{2}+(CD){}^{2}\\</p><p>=>(AC){}^{2}=(AB+BD){}^{2}+(CD){}^{2}\\</p><p>=>(AC){}^{2}=(AB){}^{2}+(BD){}^{2}+(CD){}^{2}+2.(AB).(BD)\\</p><p>=>(AC){}^{2}=(AB){}^{2}+(BC){}^{2}+2.(AB).(BD)\\</p><p>as\:\:(BD){}^{2}+(CD){}^{2}=(BC){}^{2}\\</p><p>=>(AC){}^{2}=(AB){}^{2}+(BC){}^{2}+2.(AB).(BC \:cos(theta))\\</p><p>as\:\:BD=BC \:cos(theta)\\</p><p>=>(R){}^{2}=(P){}^{2}+(Q){}^{2}+2.P.Q.cos(theta)\\</p><p>=>R=\sqrt{(P){}^{2}+(Q){}^{2}+2.P.Q.cos(theta)}</p><p>$

$\large\mathcal{\bf{\boxed{{\large\mathcal{R=\sqrt{(P){}^{2}+(Q){}^{2}+2.P.Q.cos(theta)}}}}}}</p><p>$

$Now\:\: \\ </p><p>tan(phie)=\frac{CD}{AD}\\</p><p>=>tan(phie)=\frac{CD}{AB+BD}\\</p><p>=>tan(phie)=\frac{BC\: sin(theta)}{AB+BC\:cos(theta)}$

$\huge\mathcal{\bf{\boxed{{\large\mathcal{tan(phie)=\frac{Q\: sin(theta)}{P+Q\:cos(theta)}}}}}}$

‼️✓✓Special cases:

1)

If AB and BC vectors are perpendicular to each other...i.e.,(theta)=90°

Therefore....

tan(phie)=Q/P

And

R²=P²+Q²

(2)

If the vectors are directed in opposite direction to each other...i.e.(theta)=180°

I.e..cos(180°)=-1

Now ...

R²=P²+Q²+2PQ(-1)=(P-Q)²

=>R=P-Q (this is the minimum resultant)

R=P-Q (this is the minimum resultant)And......

tan(phie)=0

=>(Phie)=0°

(3)

If the vectors are in the same direction then...(theta)=0°

Now ...cos(0°)=1

R²=P²+Q²+2(P)(Q)=(P+Q)²

=>R=(P+Q) (this is the maximum resultant)

R=(P+Q) (this is the maximum resultant)And

tan(phie)=0

=>(Phie)=0°

➡️➡️✓✓important::::

Now ..if |P|=|Q|

Then .....

R²=P²+P²+2(P)(P)cos(theta)

=>R²=2P²(1+cos(theta))

=>R²=4P²cos²(theta/2)

=>R=2Pcos(theta/2)

Or...

=>R=2Qcos(theta/2)

And

$tan(phie)=\frac{Q\:sin(theta)}{P+Q\:cos(theta)}\\</p><p>=>tan(phie)=\frac{sin(theta)}{1+cos(theta)}\\</p><p>=>tan(phie)=\frac{2\:sin(theta/2)\:cos(theta)}{2\:cos {}^{2} (theta/2)} \: \\ = > tan(phie)=\frac{sin(theta/2)}{cos(theta/2)}\\ = > tan(phie)=tan(theta/2) \\ </p><p>=>phie=theta/2$

$\large\mathcal{\bf{\boxed{\large\mathcal{phie=\frac{theta}{2}}}}}</p><p>$

$\large\mathcal\red{hope\: this \: helps \:you......Dorri......}$

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