Acircuit consists of a power supply and two resistors r1 and r2 connected in parallel. an ammeter is placed to measure the current through the restive r1. the ammeter reads i1= 7.7ma and the values of the resistors are r1=6.0 k ohm and r2= 1.6 ohm. the power supply was turned off, then the circuit was modified. resistor r1 was replaced with a different resistor rnew= 1.2 ohm. the power supply is turned back on. it provides exactly the same voltage as originally. what is the new current through resistor rnew?

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Ответ:

mmm119

21.03.2022

Physics

$5.2\ \text{m/s}$

$70^{\circ}$ south of east

Explanation:

v_a = 3 m/s

$\theta_a$ = $20^{\circ}$ north of east

v_b = 6 m/s

$\theta_b$ = $40^{\circ}$ south of east = $360-40=320^{\circ}$ north of east

x and y component of v_a

$v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}$

$v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}$

x and y component of v_b

$v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}$

$v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}$

$\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}$

Magnitude

$|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}$

Direction

$\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}$

The magnitude of the change in velocity vector is $5.2\ \text{m/s}$ and the direction is $70^{\circ}$ south of east.

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