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insens350 [35]
2 years ago
15

HELP PLEASE I REALLY NEED THE ANSWERS​

Physics
1 answer:
Paul [167]2 years ago
8 0

\star\:{\underline{\underline{\sf{\purple{ \: Question \: 19\: }}}}}

{\large{\textsf{\textbf{\underline{\underline{Given \: :}}}}}}

☆ Number of resistors = 5

☆ Resistance of each resistors = \tt \dfrac{1}{5} \text{\O}mega

{\large{\textsf{\textbf{\underline{\underline{To \: Find \: :}}}}}}

☆ Maximum resistance which can be made.

{\large{\textsf{\textbf{\underline{\underline{Solution \: :}}}}}}

• To get maximum resistance, connect the all the given resistances in series.

For,

<u>Series combination :- </u>

\longrightarrow  \sf  R_{s} = R_{1} + R_{2} + R_{3} + R_{4} + R_{5}

\longrightarrow  \sf   R_{s}= \dfrac{1}{5}  +  \dfrac{1}{5}  +  \dfrac{1}{5}  +  \dfrac{1}{5}  +  \dfrac{1}{5}

\longrightarrow  \sf  R_{s} = \dfrac{1 + 1 + 1 + 1 + 1}{5}

\longrightarrow  \sf R_{s}  = \cancel{ \dfrac{5}{5}  }

\longrightarrow  \sf R_{s}  =  \purple{1  \: \text{\O}mega}

\therefore <u>Maximum resistance is </u>1 Ω.

\star\:{\underline{\underline{\sf{\red{ \: Question \: 20\: }}}}}

{\large{\textsf{\textbf{\underline{\underline{Given \: :}}}}}}

☆ Number of resistors = 5

☆ Resistance of each resistors = \tt \dfrac{1}{5} \text{\O}mega

{\large{\textsf{\textbf{\underline{\underline{To \: Find \: :}}}}}}

☆ Minimum resistance which can be made.

{\large{\textsf{\textbf{\underline{\underline{Solution \: :}}}}}}

• To get minimum resistance, connect the all the given resistances in parallels.

For,

<u>Parallel combination :- </u>

\longrightarrow \sf \dfrac{1}{R_{p}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}

\longrightarrow \sf\dfrac{1}{R_{p}}= \dfrac{1}{ \frac{1}{5} }  + \dfrac{1}{ \frac{1}{5} }  + \dfrac{1}{ \frac{1}{5} }  + \dfrac{1}{ \frac{1}{5} }  + \dfrac{1}{ \frac{1}{5} }

\longrightarrow \sf\dfrac{1}{R_{p}}= 1 \times  \dfrac{5}{1}   +1 \times  \dfrac{5}{1}   +1 \times  \dfrac{5}{1} + 1 \times  \dfrac{5}{1} + 1 \times  \dfrac{5}{1}

\longrightarrow \sf\dfrac{1}{R_{p}}=  \dfrac{5}{1}   +  \dfrac{5}{1}   +   \dfrac{5}{1} +  \dfrac{5}{1} +   \dfrac{5}{1}

\longrightarrow \sf\dfrac{1}{R_{p}}=  25

\longrightarrow \sf R_{p}=   \red{ \dfrac{1}{25}  \: \text{\O}mega}

\therefore <u>Minimum resistance is</u> \dfrac{1}{25} Ω.

{\large{\textsf{\textbf{\underline{\underline{Note \: :}}}}}}

☆ Figure in attachment.

{\underline{\rule{290pt}{2pt}}}

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A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
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A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

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Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

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3 years ago
PLEASE ANSWER FAST!!! WILL GIVE BRAINLIEST!!!!!
Alex17521 [72]
I would go with Segment D.
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3 years ago
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