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3 years ago

Which of the following requirement statements is an example of a breakdown of the accuracy standard?

Engineering

1 answer:

const2013 [10]3 years ago

4 0

Answer:

<u>The automobile rental prices shall show all taxes (including a 6% state tax).</u>

Explanation:

Im pretty sure

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A battery weighing 0.525 lb can store 30.8 kcal of electrical energy. What is the energy density of the battery in Wh per kg? Gi

MatroZZZ [7]

Answer:

Energy Density: 129.412 $\frac{Wh}{kg}$

Explanation:

First of all, let's convert the battery weighing from pounds to kilograms:

0.525 lb x $(\frac{0.454 kg}{1 lb})$ = 0.238 kg

Now let's convert the electrical energy from kcal to Wh:

30.8 kcal x $(\frac{1 Wh}{1 kcal})$ = 30.8 Wh

Now we can calculate energy density as follows:

Energy Density = $\frac{30.8}{0.238}$

Energy Density = 129.412 $\frac{Wh}{kg}$

4 0

4 years ago

A conducting spherical shell having an inner radius of 3 cm and outer radius of 4.3 cm carries a net charge of 9.6 μC. A conduct

Arte-miy333 [17]

Answer:

sigma = 3.4*10^-4 C / m^2

Explanation:

Given:

- The net charge of the conducting sphere Q_s = 1.7 uC

- The net charge in the shell Q_shell = 9.6 uC

- Outer radius of the shell r = 4.3 cm

Find:

Determine the surface charge density on the outer surface of the shell.

Solution:

- The concept here is that the net charge on the conducting sphere must be equal to charge on inner shell. Hence, Q_s = Q_in

Q_shell = Q_in + Q_out

Q_out = Q_shell - Q_in

Q_out = 9.6 - 1.7 = 7.9 uC

- Now we can calculate the charge density of the outer shell by:

sigma = Q_out / A_out

sigma = (7.9*10^-6) / 4*pi*(0.043)^2

sigma = 3.4*10^-4 C / m^2

4 0

3 years ago

A cylindrical specimen of a cold-worked brass has a ductility (%EL) of 27%. If its cold-worked radius is 14 mm, what was its rad

blagie [28]

Answer:

Initial radius of cylindrical specimen

ri = 15.46mm

Explanation:

Cw = [(A1 - A2)/A1] * 100% ...equation 1

Where CW = cold worked percentage gain for( 27% EL) ductile

A2 = final area of cylindrical specimen = pai * r²

Where r = final radius of cylindrical specimen

So therefore expanding equation 1 will give

A1 * Cw = (100 * A1) - (100 * A2)

Cw = 18% from attached graph

A2 = area of cylindrical specimen after cold work = pai * 14² = 615.75mm²

So therefore

18A1 = 100A1 - (100 * 615.75)

-82A1 = -61575.22

A1 = 750.92mm²

So solving for initial radius from initial area of specimen

r1 = (A1/pai)^½

r1 = (750.92/pai)^½

r1= 15.46mm

5 0

3 years ago

Kkhghghglgghklghghghlk

RideAnS [48]

Answer:

hLDskjdbKSABCLABJC

Explanation:

BECAUSE

3 0

4 years ago

Read 2 more answers

Lawn maintenance is an alternative energy source<br> -true<br> -false

Reika [66]

Answer:

false

Explanation:

4 0

3 years ago