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bearhunter [10]
2 years ago
11

When different resistors are connected in parallel across an ideal battery, we can be certain that: a) their equivalent resistan

ce is equal to the average of the individual resistances. b) the potential difference across each is the same. c) the power dissipated in each is the same. d) the same current flows in each one. e) their equivalent resistance is greater than the resistance of any one of the individual resistances.
Physics
1 answer:
sergeinik [125]2 years ago
5 0

Answer:

b) the potential difference across each is the same.

Explanation:

When resistors are connected in parallel with each other

then the terminals of all the resistors will connected across the terminals of the battery

So we know the potential difference of the battery is same across all the resistors

So we can say that the equivalent resistance of all the resistance is

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

so its not the mean of all resistors

also we know that the resistance are all different so power across each resistance is different given as

P = \frac{V^2}{R}

also current in each resistance is also different and given by

i = \frac{V}{R}

so correct answer will be

b) the potential difference across each is the same.

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Answer:

819.78 m

Explanation:

<u>Given:</u>

  • OA = range of initial position of the airplane from the point of observation = 375 m
  • OB = range of the final position of the airplane from the point of observation = 797 m
  • \theta = angle of the initial position vector from the observation point = 43^\circ
  • \alpha = angle of the final position vector from the observation point = 123^\circ
  • \vec{AB} = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} =  (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} =  (-708.34\ \hat{i}+412.67\ \hat{j})\ m

The magnitude of the displacement vector = \sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

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