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3 years ago

What term best describes the geologic event taking place in the above illustration?

Physics

2 answers:

grigory [225]3 years ago

8 0

u lying you made me get it wrong, for ya'll out there who want the real answer is sea floor spreading

Sunny_sXe [5.5K]3 years ago

6 0

Answer:

seafloor spreading

Explanation:

For study island

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Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun

AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561 × 10¹⁹ electrons per second = 1.561 × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0

3 years ago

If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict

Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

$\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$

R = 0.0821 L atm/ K mol

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b = 0.03219 L / mol

So,

$\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15$

$P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}$

$P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}$

⇒P (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

4 0

3 years ago

A car initially traveling at 17.1 mph comes to rest in 9.7s what was its acceleration in this time?

ra1l [238]

Answer:

$a=-.78m/s^2$

Explanation:

Δ $v=at$

Δ $v$ is the difference in velocity before and after a given time.
$a$ is the acceleration of the object during this time.
$t$ is time

$(v_f-v_i)=at$ is another way to write this equation.

The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δ $v=(v_f-v_i)$

$v_f-v_i=at\\\frac{at}{t}=\frac{v_f-v_i}{t}\\a=\frac{v_f-v_i}{t}$

I rearranged this equation to solve for $a$ , but this is a step that you don't need to take, it's just good to get in the habit of doing this.
Plug in the given values. Note that our final velocity is $0$ , because the car travels until at rest.

$a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}$

Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.

$a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}$

if you converted correctly, your answer for $v_f$ will be ≅ $7.6m/s$ .
Now divide. Notice that the units for acceleration are $m/s^2$ or meters per second, per second.

$a=\frac{-7.6m/s}{9.7s}\\a=-.78m/s^2$

Our final answer is negative because the car is slowing down. Do not square this answer as the square symbol only applies to the units, not the magnitude.

4 0

3 years ago

Read 2 more answers

Suppose the Sun appeared to you 900 times dimmer than it does now. How far away from the Sun would you be? (A) 1/9 AU (B) 3 AU

Harrizon [31]

Answer:

The correct answer is option 'c': 30 AUs

Explanation:

For a spherical wave front emitted by sun with total energy 'E' the energy density over the surface when it is at a distance 'r' from the sun is given by

$e=\frac{E}{4\pi r^{2}}$

This energy per unit area is sensed by observer as intensity of the sun.

Let the initial intensity of sun at a distance $r_{1}$ be $e_{1}$

Thus if the sun becomes 900 times dimmer we have

$e'=\frac{e_{1}}{900}\\\\\frac{E}{4\pi r_{2}^{2}}=\frac{1}{900}\times \frac{E}{4\pi r_{1}^{2}}\\\\\Rightarrow r_{2}^{2}={r_{1}^{2}}\times 900\\\\\therefore r_{2}={r_{1}}\times {30}$

Thus the distance increases 30 times.

5 0

3 years ago

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive rea

Slav-nsk [51]

Answer:

the capacitive reactance is reduced by a factor 2

Explanation:

The capacitive reactance is given by:

$X_C = \frac{1}{2\pi f C}$

where

f is the frequency of the source

C is the capacitance

In this case, the frequency of the source is doubled:

f' = 2f

while the capacitance does not change. So the new capacitive reactance will be

$X_C' = \frac{1}{2\pi f' C}=\frac{1}{2\pi (2f)C}=\frac{1}{2}X_C$

so the capacitive reactance is reduced by a factor 2.

5 0

3 years ago