Complete Question
The complete question is shown on the first uploaded image
Answer:
The maximum emf is 
The emf induced at t = 1.00 s is 
The maximum rate of change of magnetic flux is 
Explanation:
From the question we are told that
The number of turns is N = 44 turns
The length of the coil is 
The width of the coil is 
The magnetic field is 
The angular speed is 
Generally the induced emf is mathematically represented as

Where
is the maximum induced emf and this is mathematically represented as

Where
is the magnetic flux
N is the number of turns
A is the area of the coil which is mathematically evaluated as

Substituting values


substituting values into the equation for maximum induced emf


given that the time t = 1.0sec
substituting values into the equation for induced emf 


The maximum induced emf can also be represented mathematically as

Where
is the magnetic flux and
is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

A. overgrazing of livestock
Explanation:
Overgrazing of livestock is the ranch practice that contributes to soil erosion.
Soil erosion is the washing away of the top layer of the earth on which plant grows.
- Plants are covers of the earth and they prevent agents of denudation from actively washing away the soil.
- When livestock are introduced into a place, continuous grazing leaves the surface bare.
- The plants becomes depleted and the surface is exposed.
- This fast tracks the process of overgrazing.
learn more:
Soil erosion brainly.com/question/2473244
#learnwithBrainly
The particle has acceleration vector

We're told that it starts off at the origin, so that its position vector at
is

and that it has an initial velocity of 12 m/s in the positive
direction, or equivalently its initial velocity vector is

To find the velocity vector for the particle at time
, we integrate the acceleration vector:

![\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20v%3D%5Cleft%5B12%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cjmath)
![\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20v%3D%5Cleft%5B12%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5C%2C%5Cvec%5Cjmath)
Then we integrate this to find the position vector at time
:

![\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%2812%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cjmath)
![\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cleft%5B%5Cleft%2812%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29t%2B%5Cleft%28-1.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%282.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5C%2C%5Cvec%5Cjmath)
Solve for the time when the
coordinate is 18 m:

At this point, the
coordinate is

so the answer is C.