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2 years ago

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,

Series combination :- 

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

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

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

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

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

$\therefore$ Maximum resistance is 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,

Parallel combination :- 

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

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

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

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

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

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

$\therefore$ Minimum resistance is $\dfrac{1}{25}$ Ω.

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

☆ Figure in attachment.

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

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Read 2 more answers

A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points

Harman [31]

Answer:

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

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3 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Then the electric field at the point of interest is estimated as:

$E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}$

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3 years ago

HELP PLEASE I REALLY NEED THE ANSWERS​

HELP PLEASE I REALLY NEED THE ANSWERS