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kramer
3 years ago
12

The magnetic flux through a loop:

Physics
2 answers:
dexar [7]3 years ago
4 0

Answer:

The magnetic flux through a loop is zero when the B field is perpendicular to the plane of the loop.

Explanation:

Magnetic flux are also known as the magnetic line of force surrounding a bar magnetic in a magnetic field.

It is expressed mathematically as

Φ = B A cos(θ) where

Φ is the magnetic flux

B is the magnetic field strength

A is the area

θ is the angle that the magnetic field make with the plane of the loop

If B is acting perpendicular to the plane of the loop, this means that θ = 90°

Magnetic flux Φ = BA cos90°

Since cos90° = 0

Φ = BA ×0

Φ = 0

This shows that the magnetic flux is zero when the magnetic field strength B is perpendicular to the plane of the loop.

scoundrel [369]3 years ago
4 0

Answer:

D. depends just on the B field going through the loop

Explanation:

The magnetic flux trough a loop is given by the formula:

\Phi_B=\vec{B}\cdot \vec{A}

\Phi_B=BAcos\theta

Where B is the magnitude of the magnetic field and A is the area of the loop.

It is clear that the flux is maximum when the angle between the direction of B and the direction of the normal vector of A is zero (cos0 = 1).

Furthermore, we can notice that only the magnetic field that crosses the loop contributes to the flux.

Hence, the correct answer is:

D. depends just on the B field going through the loop

hope this helps!!

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Here is the answer to the given question above. If Angela has been feeling fatigued and a test is used to check the basal metabolic rate and revealed that she has a low metabolic rate, therefore, the possible diagnosis for Angela would be HYPOTHYROIDISM. <span>The BMR test works by precisely measuring the amount of oxygen that you consume when your body is basal, or completely at rest. Hope this answers your question.</span>
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4 years ago
An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
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Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
  • T_3 = The third temperature of air inside the tire = 85^\circ C =(273+85)\ K=358 \ K
  • V_1 = The first volume of air inside the tire
  • V_2 = The second volume of air inside the tire = 30\% V_1 = 0.3V_1
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)

Part (a):

Using the above equation for this part of compression in the air, we have

\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa

Hence, the pressure in the tire after the compression is 3.81\times 10^5\ Pa.

Part (b):

Again using the equation for this part for the air, we have

\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

8 0
3 years ago
Suppose your local apparent solar time is 11 am, and you have a clock that says it is noon in Greenwich. What is your longitude
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The longitude of a complete rotation of the earth in a 24 hours is calculated  as follows;

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The time difference between the local apparent solar time and the Greenwich time is calculated as follows;

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Since it is one hour time difference, the longitude is 15 degrees.

Learn more about Earth longitude here: brainly.com/question/1939015

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d_1 = 16(\frac{1}{2}gt^2)

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3 years ago
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