Which items are NOT found on a

Suppose that the cost of electrical energy is $0.15 per kilowatt hour and that your electrical bill for 30 days is $80. Assume t

mr_godi [17]

Answer:

a) 740 W b) 6.2 A c) 8.1%

Explanation:

We need first to get the total energy spent during the 30 days, that can be calculated as follows:

1 month = 30 days = 720 hr

If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:

80 $/mo = 0.15 $/Kwh*x Kwh/mo

Solving for the total energy spent in the month:

$x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh$

Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:

$P = \frac{E}{t} = \frac{533.3 kWh}{720 h} = 0.74 kW = 740 W$

b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:

$I =\frac{P}{V} =\frac{740W}{120V} = 6.2A$

c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:

P = 740 W - 60 W = 680 W

The new value of the energy spent during the entire month will be as follows:

E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo

The reduction in percentage regarding the total energy spent can be calculated as follows:

ΔE = $\frac{533.3-490}{533.3} = 0.081*100= 8.1%$

⇒ ΔE(%) = 8.1%

$%(E) =\frac{533.3-490}{533.3} = 0.081 * 100 = 8.1%$

4 0

4 years ago

In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri

ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

Here

$\sigma_x$ is the normal stress which is 1750 psf
$\sigma_y$ is 0
$\tau_{xy}$ is the shear stress which is 800 psf

So the formula becomes

$\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}$

Similarly, the minimum normal stress is given as

$\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}$

The maximum shear stress is given as

$\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}$

5 0

3 years ago

Inhalation is the most common way for a(n)

Mademuasel [1]

inhalation is the most common way for chemical to enter the body

6 0

3 years ago

What's the best way to find the load capacity of a crane? Select the best option. Call the manufacturer Ask co-workers Look at t

Orlov [11]

Answer: You need to lift a load of 15 tons (30,000 pounds) a distance of 25 feet. The distance is measured from the center pin of the crane to the center of the load. Once you determine the distance, look on that line for the largest capacity; that will indicate how many feet of boom must be extended

Explanation:

5 0

3 years ago

Question 2) A material that is malleable can be defined as:

Reil [10]

Answer:

A.

Explanation:

Able to return to it's original shape after distortion.

7 0

3 years ago