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Elenna [48]
3 years ago
5

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

ic field. A field of 1.7T is directed along the positive z-direction, which is upward.
Part A: If the loop is removed from the field region in a time interval of 2.4

ms , find the average emf that will be induced in the wire loop during the extraction process.



Part B:If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?



Could you please explain this in a step by step process, and include reasoning to every assumption in relation to the Right Hand Rule? Thanks.
Physics
1 answer:
Drupady [299]3 years ago
3 0

Answer:

See answer

Explanation:

The area of the circular loop is given by:

A = \pi r^2

The magnetic flux is given by:

\phi = \int \vec{B} \cdot d\vec{A}

d\vec{A} is parallel to \vec{B} and \vec{B} is constant in magnitude and direction therefore:

\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2

Part A)

initially the flux is \phi =\pi B r^2

after the interval \Delta t= 2.4 [m/s]

the flux is

\phi = 0

now, the EMF is defined as:

\epsilon =- \frac{d \phi}{dt},

if we consider \Delta t= 2.4 [m/s] very small then we can re-write it as:

\epsilon =- \frac{\Delta \phi}{\Delta t}

\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12

then:

\epsilon =- \frac{-0.12}{0.0024} = 50 [V]

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

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jekas [21]

Answer:

a)    Fa = G m2 [M / r² - m / (d-r)²]  

b) r2 = 31 10⁶ m

Explanation:

The equation of the law of universal gravitation is

            F = G m1 m1/ r²

The value of the gravitation constant is 6.67 10-11 N m²/kg². This force is always attractive.

Let's calculate the value of that force on the spacecraft, add the strength

Earth's force to the ship

            F1 = G M m2 / r²

The moon force ship

            F2 = G m m2 / (d-r)²

Total force is

            Fa = F1 — F2

            Fa = G M M2 / r² - G m m2 / (d-r)²

            Fa = G m2 [M / r² - m / (d-r)²]

This is the force on the spaceship

b)  Let's look for the point where the force is zero, for this we can see that the value of the bracket must be zero

           Fa = 0

           [M / r² - m / (d-r)²] = 0

            M / r² = m / (d-r)²

           (d-r)² = m/M   r²

           d² -2rd + r² - m/M   r² = 0

           r² [m/M - 1] + r 2d - d² = 0

This is a second degree equation for r, we solve the to find the results.

           r = {-2d ±√[4d² - 4 [m/M -1] (-d²)]} / (2 [m/M-1])

           r = {-2d ± √ [4d² (1 + (m/M-1)]} / 2(m/M-1)

           r = {-2d ± 2d √(m/M)}  / (2(m/M-1))

           r = 2d {-1 ± √(m/M)} / 2(m/M-1)

           r = d [-1 ± √(m/M)] /  (m/M-1)

To find the explicit value we substitute the values ​​that we can find in tables

          m = 7.36 1022 kg

          M = 5.98 1024 kg

          d = 380000 km (1000m / 1 km) = 380 10⁶ m

          r = 380 10⁶ [-1 ±√(7.36 10²² / 5.98 10²⁴)] /(7.36 10²² / 5.98 / 10²⁴ -1)

          r = 380 10⁶ [-1 - √ (1.23 10²)] / (123-1)

          r = 380 106 [-1 ± 11] / 122

 

          r1 = 380 10⁶ 10/122 = 380 10⁶ 0.08197

          r1 =  31 10⁶ m

          r2 = 380 106 [-12/122] = 380 10 6 0.09836

          r2 = -37 10⁶ m

The correct distance is the positive r2 = 31 10⁶ m

c) let's use Newton's second law, to find the acceleration in the spacecraft

 

          F = m a

          a = Fa / m2 = G m2 [M / r² - m / (d-r)²] / m2

          a = G [M/r² - m/(d-r)²]

Since we have acceleration, we can use the definition of kinematics

           a = dv / dt = dv / dr dr / dt = dv / dr v

           v dv = a dr

           v dV = G [M /r² - m /(d-r)²] dr

We integrate

            ½ (V² - Vo²) = G [M (-1 /r) -m (1 / (d-r)

We evaluate between the initial point where we can assume that the initial velocity is zero for the Xo position and the final point with velocity v at the points

            V² = 2G [M (1 / Xo - 1 /X) - m (1 / (d-X) - 1 /(d-xo)]

            V² = [-M /X -m /(d-X)] 2G + constant

We now use the definition of speed

            v = dx / dt

            dx = V dt

We substitute, perform the integral and simplify, if we can make the constant zero

             dx = √([-M / X -m / (d-X)] 2G) dt

             dx / √([-M / X -m / (d-X)] 2G = dt

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3 years ago
Does the number of particles change as the substance changes its state?​
Leviafan [203]

Particles stay the same unless there is a chemical change whether the matter is solid, liquid or gas. ... When substances change state there is no change in mass so if 100 g of ice is melted 100g of water are formed this will boil to form 100g of steam (this is called "conservation of mass").

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3 years ago
A student made the following observations about a squirrel’s movements:
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The correct option is <u>D</u>.

Qualitative observations are observations that are made using our senses of sight, hearing, smell, taste and feel. These observations do not involve numbers or measurements of any kind.

The student's observations regarding the squirrel as is mentioned in options A, B and C involve measurements. Therefore these are not qualitative observations.

Option D, however, is made on the basis of sight, where the student observes the squirrel moving in a zigzag manner.

Therefore, of all the three observations, the student's observation that the squirrel ran in a zigzag pattern is the qualitative observation.

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You drop your cell phone. Prior to hitting the ground, the phone's kinetic energy will ________ and its potential energy will __
SVETLANKA909090 [29]

Answer:

The kinetic energy of the phone would increase. The gravitational potential energy of the phone would decrease.

Explanation:

The kinetic energy {\rm KE} of an object is proportional to the square of the speed of that object. If air resistance is negligible, the phone would accelerate under gravitational pull and speed up. Hence, the kinetic energy of the phone would increase.

The gravitational field near the surface of the earth is approximately constant. Hence, the gravitational potential energy {\rm GPE} of the phone would be proportional to its height. As the phone approaches the ground, the height of the phone becomes lower and the gravitational potential energy of the phone would decrease.

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1 year ago
an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity
MatroZZZ [7]
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\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Displacement}{Total\:Time}

\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}

\\ \rm\longrightarrow Avg\:Velocity= 17m/s

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3 years ago
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