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3 years ago

The resistor if resistance 50N is connected to a potential difference of 10V

1 answer:

7 0

1. V = IR
You rearrange it to make I the subject by dividing both sides by R. You get:
I = V / R
I = 10 / 50
= 0.2 A

2. Q = It
Charge = 0.2 x 2
Charge = 0.4 C

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If the lily pads are spaced 2.4 m apart and Ferdinand jumps with a speed of 5 m/s taking 0.6 s to go from lily pad to lily pad,

Zina [86]

Answer:

$36.87^{\circ}$

Explanation:

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Horizontal range is given by

$R=v_xT\\\Rightarrow R=vcos\theta T\\\Rightarrow cos\theta=\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{2.4}{5\times 0.6}\\\Rightarrow \theta=36.87^{\circ}$

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3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

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3 years ago