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4 years ago

What is the charge value of 45 electrons

Physics

1 answer:

tamaranim1 [39]4 years ago

8 0

Answer:

$7.2\times 10^{-18}C$

Explanation:

-Let x be the charge of 45 electrons

-An electron has a relatively law mas.

-Using the law of proportions, the charge value of 45 electrons is calculated as:

$1e=1.6 \times 10^{-19} coulomb\\45e=x\\\\x=\frac{45\times1.6 \times 10^{-19} coulomb}{1e}\\\\x=7.2\times 10^{-18}C$

Hence, the electron charge of 45 electrons is $7.2\times 10^{-18}C$

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A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap

viktelen [127]

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

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3 years ago

You normally drive a 12-h trip at an average speed of 100 km/h . Today you are in a hurry. During the first two-thirds of the di

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Answer:

78 km/h

Explanation:

If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:

100 km/h · 12 h = 1,200 km

Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.

1,200 * 2/3 = 800 km

I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.

To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.

116 km/1 h = 800 km/? h
800 = 116 · ?
? = 800/116
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I spent 6.89655172 hours driving during the first 2/3 of the distance.

Now, I need to subtract this value from 12 hours to find the remaining time I have left.

12 h - 6.89655172 h = 5.10344828 h

Using this remaining time and my remaining distance, I can calculate my average speed.

? km/1 hr = 400 km/5.10344828 h
5.10344828 · ? = 400
? = 400/5.10344828
? = 78.3783783148

My average speed during the last third of the distance is around 78 km/h.

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