Ask question

Ask question

All categories

3 years ago

14

The current in a wire is 4.00A. How much time is needed for 1 mole of electrons (6.02×1023 electrons) to pass a point in the wir

e?

1 answer:

scZoUnD [109]3 years ago

8 0

Answer:

t= 24080 s

Explanation:

Given that

Current in the wire ,I = 4 A

The charge ,q = 6.02 x 10²³ e C

We know that

$I=\dfrac{q}{t}$

I=Current

q=Charge

t=time

$t=\dfrac{q}{I}$

Now by putting the values in the above equation we get'

$t=\dfrac{6.02\times 10^{23}\times 1.6\times 10^{-19}}{4}\ s$

t= 24080 s

You might be interested in

A 9.5 v battery supplies a 3.5 ma current to a circuit for 6.0 h . part a how much charge has been transferred from the negative

dmitriy555 [2]

The total time is (1h=3600 s):
$\Delta t=6.0 h=21600 s$
The current intensity I is the amount of charge Q that passes through a certain point in a time $\Delta t$ :
$I= \frac{Q}{\Delta t}$
Since we know the current, $I=3.5 mA=0.0035 A$ , we can find how much charge has been trasferred from one terminal to the other in 6 hours:
$Q= I \Delta t=(0.0035 A)(21600 s)=75.6 C$

3 0

3 years ago

Which will settle out if allowed to sit on a table? suspension colloid solution compound a suspension a colloid a solution a com

liubo4ka [24]

Answer is suspension.

Lets define all options.

<h3>Suspension:</h3>

In suspension the solute does not dissolve in liquid. When placed on table for some time, it will settle down at the bottom of the beaker. We can separate particles of solute easily from solvent through filtration.

<h3>Colloid:</h3>

In colloid particles of solute does not dissolve in liquid neither it is settle down. It floats through the solvent. It cannot be separated by filtration.

<h3>Solution:</h3>

In solution the particles of solute dissolve in to the solvent. We cannot identify them as separate. We cannot separate them by filtration. Salt and water solution is an example of it. Evaporation is the technique that is required to separate them.

<h3>Compound:</h3>

In compound, the two elements combine to form a new thing. Resultant/ compound have new or different properties other than its ingredients.

Now, the question was which of the following allow to settle out when sit on a table, so the answer is suspension. Suspension allows the particles to settle out when sit on a tables for some time.

8 0

4 years ago

Read 2 more answers

Halp meeeeeeeeeeeeeeeeee

Greeley [361]

Answer:

1º all

2º 1

3º 2

4º the same

Explanation:

7 0

3 years ago

A fighter plane flying at constant speed 420 m/s and constant altitude 3300 m makes a turn of curvature radius 11000 m. On the g

Arada [10]

Answer:

"Apparent weight during the "plan's turn" is 519.4 N

Explanation:

The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is $\mathrm{a}_{\mathrm{n}}=\frac{v^{2}}{R}$

Given that,

v = 420 m/s

R = 11000 m

Substitute the values in the above equation,

$a_{n}=\frac{420^{2}}{11000}$

$a_{n}=\frac{176400}{11000}$

$a_{n}=16.03 \mathrm{m} / \mathrm{s}^{2}$

It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vector $W_{\mathrm{app}}=m(\text { vector } g \text { -vector a })$

In magnitude,

$| \text { vector } g-\text { vector } a |=\sqrt{\left(g^{2}+a^{2}\right)}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{\left(9.8^{2}+16.03^{2}\right)}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{(96.04+256.96)}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{353}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=18.78 \mathrm{m} / \mathrm{s}^{2}$

Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

$\mathrm{W}_{\mathrm{app}}=(18.78 \mathrm{m} / \mathrm{s}^ 2)(53 \mathrm{kg})=995.77 \mathrm{N}$

Which is quite heavier than his/her true weigh of 519.4 N

7 0

4 years ago

Read 2 more answers

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

3 years ago