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2 years ago

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How many electrons pass a given point in the circuit in 3 min? The fundamental charge is 1.602 × 10−19 C. amps: 0.415 , volts: 1

7 , ohms: 41

1 answer:

morpeh [17]2 years ago

4 0

The no of electron passing through the circuit is current by time. The no of electron passing are 46.63 x 10¹⁹ electrons.

<h3>What is charge?</h3>

The charge is the physical quantity which attracts or repels another object when comes into its field.

The time 3 min has 3 x 60 =180 seconds.

Total Charge Q = current I x time t

Q = 0.415 x 180

Q =74.7 coulombs

The no of electrons is the ratio of total charge divided by the fundamental charge.

n = 74.7/ 1.602 × 10⁻¹⁹

n= 46.63 x 10¹⁹ electrons

Thus, electrons pass a given point in the circuit in 3 min are 46.63 x 10¹⁹ .

Learn more about charge.

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3 years ago

You are climbing in the high sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. to find the height of t

sergey [27]

The total time it takes the sound to reach our on top of the cliff (10.0 s) is actually sum of two times:
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1) The rock moves by uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ . Its law of motion is given by $y(t)=h-v_0t-\frac{1}{2}gt^2$ , where $v_0=0$ is the initial velocity of the rock, and h is the height of the cliff. The time t1 is the time at which the rock reaches the ground, so that y(t1)=0, and the equation becomes
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2) The sound moves from the ground to the top of the cliff by uniform motion, with constant speed v=343 m/s. Therefore, the sound covers the distance h (the height of the cliff) in a time t2 given by
$h=vt_2$
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$=\frac{1}{2}gt_1^2=vt_2$ (1)
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3 0

3 years ago

Read 2 more answers

A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t

Mice21 [21]

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}at^2$

$a=\dfrac{2s}{t^2}$

$a=\dfrac{2\times 12\ m}{(1.2\ s)^2}$

a = 16.67 m/s²

Now put the value of a in equation (1) as :

$q=\dfrac{ma}{E}$

$q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}$

q = 0.0000249 C

or

$q=2.49\times 10^{-5}\ C$

Hence, this is the required solution.

5 0

3 years ago

A carbon rod with a radius of 1.9 mm is used to make a resistor. What length of the carbon rod should be used to make a 3.7 Ω re

vladimir2022 [97]

Answer:

Length = 2.32 m

Explanation:

Let the length required be 'L'.

Given:

Resistance of the resistor (R) = 3.7 Ω

Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]

Resistivity of the material of rod (ρ) = $1.8\times 10^{-5}\ \Omega\cdot m$

First, let us find the area of the circular rod.

Area is given as:

$A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2$

Now, the resistance of the material is given by the formula:

$R=\rho( \frac{L}{A})$

Express this in terms of 'L'. This gives,

$\rho\times L=R\times A\\\\L=\frac{R\times A}{\rho}$

Now, plug in the given values and solve for length 'L'. This gives,

$L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m$

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.

5 0

3 years ago

Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5

schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0

3 years ago