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AveGali [126]
3 years ago
10

The average age of engineering students at graduation is a little over 23 years. This means that the working career of most engi

neers is almost exactly 500 months. How much would an engineer need to save each month to accrue $5 million by the end of her working career? Assume a 9% interest rate, compounded monthly.
Engineering
1 answer:
RideAnS [48]3 years ago
6 0

Answer:

$916

Explanation:

To solve this, we use the formula

FV = P/i * [(1+i)^n - 1], where

FV = future value of the all the money invested, $5 million

n = time span, = 500 months

P = payment per month

I = interest rate, 9% by 12 months, = 0.0075

Considering that we have been given all in our question, then we substitute directly and solve. So we have,

5000000 = P/0.0075 * [(1+0.0075)^500 -1]

5000000 * 0.0075 = P * [1.0075^500 - 1]

37500 = P * [41.93 - 1]

37500 = P * 40.93

P = 37500/40.93

P = $916.20

Therefore, the engineer needs to save $916 in a month which is the accrued

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a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What
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Answer:

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Explanation:

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The formula for the rate of heat conduction is given by:

dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C

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Alexxandr [17]

Answer:

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Explanation:

We are given:

m = 1.06Kg

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COP_R = \frac {1}{(T_R/T_l) -1}

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Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

T_L = \frac{T_H}{1.2};

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From the refrigerant R -134a table at T_L = 12.4°c we have 442KPa

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