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4 years ago

The product second moment of area Ixy is found by multiplying Ix and Iy. a)True b)- False

Engineering

1 answer:

Kamila [148]4 years ago

7 0

Answer:

(b)False

Explanation:

$I_{xy}$ defined as

$I_{xy}$ = $\int \left (x\cdot y\right )dA$

Where x is the distance from centroidal x-axis

y is the distance from centroidal y-axis

dA is the elemental area.

The product of x and y can be positive or negative ,so the value of $I_{xy}$ can be positive as well as negative .

So from the above expressions we can say that the product of $I_{x},I_y$ is different from $I_{xy}$ .

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Q2) An engineer borrowed $3000 from the bank, payable in six equal end-of-year payments at 8%. The bank agreed to reduce the int

tatyana61 [14]

Answer is: $637.28; just did the math but i really don’t want to type it all out.

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4 years ago

A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?

blagie [28]

Answer:

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

4 0

3 years ago

Consider the following hypothetical scenario for Jordan Lake, NC. In a given year, the average watershed inflow to the lake is 9

dybincka [34]

Answer:

The lake can withdraw a maximum of $1.464\times 10^{10}$ cubic feet per year to provide water supply for the Triangle area.

Explanation:

The maximum amount of water that can be withdrawn from the lake is represented by the following formula:

$V = V_{in}+V_{p}-V_{e}-V_{out}$ (Eq. 1)

Where:

$V$ - Available amount of water for water supply in the Triangle area, measured in cubic feet per year.

$V_{in}$ - Inflow amount of water, measured in cubic feet per year.

$V_{out}$ - Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.

$V_{p}$ - Amount of water due to precipitation, measured in cubic feet per year.

$V_{e}$ - Amount of evaporated water, measured in cubic feet per year.

Then, we can expand this expression as follows:

$V = f_{in}\cdot \Delta t+h_{p}\cdot A_{l}-h_{e}\cdot A_{l}-f_{out}\cdot \Delta t$

$V = (f_{in}-f_{out})\cdot \Delta t +(h_{p}-h_{e})\cdot A_{l}$ (Eq. 2)

Where:

$f_{in}$ - Average watershed inflow, measured in cubic feet per second.

$f_{out}$ - Average flow to be released, measured in cubic feet per second.

$\Delta t$ - Yearly time, measured in seconds per year.

$h_{p}$ - Change in lake height due to precipitation, measured in feet per year.

$h_{e}$ - Change in lake height due to evaporation, measured in feet per year.

$A_{l}$ - Surface area of the lake, measured in square feet.

If we know that $f_{in} = 900\,\frac{ft^{3}}{s}$ , $f_{out} = 300\,\frac{ft^{3}}{s}$ , $\Delta t = 31,536,000\,\frac{second}{yr}$ , $h_{p} = 32\,\frac{in}{yr}$ , $h_{e} = 55\,\frac{in}{yr}$ and $A_{l} = 47,000\,acres$ , the available amount of water for supply purposes in the Triangle area is:

$V = \left(900\,\frac{ft^{2}}{s}-300\,\frac{ft^{3}}{s} \right)\cdot \left(31,536,000\,\frac{s}{yr} \right) +\left(32\,\frac{in}{yr}-55\,\frac{in}{yr} \right)\cdot \left(\frac{1}{12}\,\frac{ft}{in}\right)\cdot (47000\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)$ $V = 1.464\times 10^{10}\,\frac{ft^{3}}{yr}$

The lake can withdraw a maximum of $1.464\times 10^{10}$ cubic feet per year to provide water supply for the Triangle area.

5 0

3 years ago

When groups of molecules within liquid and gases move, it is called

steposvetlana [31]

Answer:

convenction is your answer

7 0

3 years ago

If you were to plot the voltage versus the current for a given circuit, what would you expect the slope of the line to be? If no

Brut [27]

Answer:

Part 1: It would be a straight line, current will be directly proportional to the voltage.

Part 2: The current would taper off and will have negligible increase after the voltage reaches a certain value. Graph attached.

Explanation:

For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.

V=I*R

where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.

In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.

7 0

4 years ago