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Gnoma [55]
2 years ago
11

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
Physics
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.

brainly.com/question/11944606

#SPJ1

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How does a virus make it into a cell and what happens to it as it tries to get to its target “The Nucleus”?
Hunter-Best [27]
Once the virus gets into the host's body, it enters the nearest cell.Once it gets to the nucleus it turns that cell into a virus-production cell and once the immune system detects those viruses it takes action.
4 0
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:
- the time it takes the rock to hit the ground starting from the top of the cliff, t1
- the time it takes the sound to reach the top of the cliff starting from the ground, t2
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
0=h-\frac{1}{2}gt_1^2
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
3) If we rewrite h in both equations, we can write:
=\frac{1}{2}gt_1^2=vt_2 (1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
t_1+t_2=10
from which we find
t_2=10-t_1
if we substitute this into eq.(1), we get
\frac{1}{2}gt_1^2+vt_1-10v=0
Numerically:
4.9t_1^2 + 343 t_1 - 3430=0
Solving the equation, we find the solution t_1=8.87 s (the other solution is negative, so it does not have physical meaning). As a consequence,
t_2 = 10-t_1 = 1.13 s
and the height of the cliff is given by
<span>h=vt_2=(343 m/s)(1.13 s)=388 m</span>
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
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