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3 years ago

I need help on the Coderz Challenge missions 3 part 3. PLEASE HELP!

Engineering

2 answers:

allsm [11]3 years ago

6 0

Answer:

the answer how you analyzs the problwm

romanna [79]3 years ago

4 0

Answer:

To access a missions, simply log on to your CoderZ account and navigate to one of the courses you are assigned, choose a pack, and then choose one of the missions!

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1. (15) A truck scale is made of a platform and four compression force sensors, one at each corner of the platform. The sensor i

Elanso [62]

Answer:

a). 139498.24 kg

b). 281.85 ohm

c). 10.2 ohm

Explanation:

Given :

Diameter, d = 22 m

Linear strain, $\epsilon$ = 3%

= 0.03

Young's modulus, E = 30 GPa

Gauge factor, k = 6.9

Gauge resistance, R = 340 Ω

a). Maximum truck weight

σ = Eε

σ = $0.03 \times 30 \times 10^9$

$\frac{P}{A} =0.03 \times 30 \times 10^9$

$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$

= 342119.44 N

For the four sensors,

Maximum weight = 4 x P

= 4 x 342119.44

= 1368477.76 N

Therefore, weight in kg is $m=\frac{W}{g}=\frac{1368477.76}{9.81}$

m = 139498.24 kg

b). Change in resistance

$k=\frac{\Delta R/R}{\Delta L/L}$

$\Delta R = k. \epsilon R$ , since $\epsilon= \Delta L/ L$

$\Delta R = 6.9 \times 0.03 \times 340$

$\Delta R = 70.38$ Ω

For 4 resistance of the sensors,

$\Delta R = 70.38 \times 4 = 281.52$ Ω

c). $k=\frac{\Delta R/R}{\epsilon}$

If linear strain,

$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$ , where k = 1

$\Delta R = \frac{\Delta L}{L} \times R$

$\Delta R = 0.03 \times 340$

$\Delta R = 10.2$ Ω

4 0

3 years ago

If in Example 1.2,q = (10 - 10e2) mC, find the current at t = 0.5 s.

NeTakaya

Answer:

pls mark me as brainliest

Explanation:

Answer: 7.36 mA

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The maximum amount of water that can be withdrawn from the lake is represented by the following formula: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

A rigid insulated tank is divided into 2 equal compartments by a thin rigid partition. One of the compartments contains air, ass

Illusion [34]

Https://www.slader.com/discussion/question/an-insulated-rigid-tank-is-divided-into-two-equal-parts-by-a-partition-initially-one-part-contains-4/

there will be the answer

6 0

3 years ago

An uncharged capacitor is connected to a resistor and a battery. Choose what happens to current, potential difference and charge

Ilya [14]

Answer:

The charge on the plates will increase with time

The potential difference across the capacitor starts with zero and then increases gradually to a maximum value

The current through the circuit starts high and then drops exponentially

Explanation:

<u>Case : An uncharged capacitor is connected to a resistor and a battery in a closed circuit.</u>

The charge on the plates will increase with time

applying this equation : Q = $Q_{0} [ 1 - e^{\frac{-t}{RC} } ]$ as the value of (t) increases the value of Q increases i.e. charge on the plates

The potential difference across the capacitor starts with zero and then increases gradually to a maximum value

applying this equation : V = $V_{0} [ 1 - e^{\frac{-t}{RC} } ]$

The current through the circuit starts high and then drops exponentially

current : I = $I_{0} e^{\frac{-t}{RC} }$

6 0

3 years ago