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3 years ago

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Sarah invested 12000 in a unit trust 5 years ago, the value of the unit trust has increased by 7% per annum for the last 3 years

. before this the price had decreased by 3% per annumCalculate the current price of the unit trust

2 answers:

miss Akunina [59]3 years ago

8 0

Answer:

The value of the fund today is $13831.72

Explanation:

Let us calculate the value at the end of each year, step by step. An annual decrease of 3% corresponds to a multiplier 1.0-0.03 = 0.97. An increase of 7% corresponds to 1.07.

Value at the end of

year 1: 12000 * 0.97

year 2: 12000 * 0.97 * 0.97

year 3: 12000 * 0.97 * 0.97 * 1.07

year 4: 12000 * 0.97 * 0.97 * 1.07 * 1.07

year 5: 12000 * 0.97 * 0.97 * 1.07 * 1.07 * 1.07 = 13831.72

The value of the fund today (end of year 5) is $13831.72

Damm [24]3 years ago

3 0

The Answer to this question is

Value fund is $13831.72

Hope this helps (;

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Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

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The x-component of the electric field at the origin = -11.74 N/C.

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<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
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Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

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The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

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$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

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