What do machines require in order to work?

The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi

Zolol [24]

Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,

1 gallon = 0.133 cubic feet (cf)

Therefore,

Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133

= 4.28 x 10˄8

Applying Mass balance i.e

Accumulation = Mass In - Mass out (Eq. 01)

Here

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation = 0.5 - 11

= - 10.5 cfs

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

= 37,800

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

= 907,200

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200

= 471 Days.

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

8 0

3 years ago

Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1

Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis = $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

3 0

3 years ago

Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature

sveta [45]

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

$\epsilon_{car}=1-\frac{T_c}{T_H}$

Where $T_c$ is the atmospheric temperature and $T_H$ is the maximum burn temperature.

For the case (B), the efficiency we will use is:

$\epsilon_{b}=0.4\epsilon_{car}$

The work done by the engine can be calculated as:

$W=\epsilon Q=\epsilon H_v\cdot m_{fuel}$ where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

$m=\frac{P\cdot t}{\epsilon H_v}$

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

$TC=m\cdot c$

8 0

3 years ago

A material has the following properties: Sut = 275 MPa and n = 0.40. Calculate its strength coefficient, K.

Tems11 [23]

Answer:

The strength coefficient is K = 591.87 MPa

Explanation:

We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by

$S_{ut}=K \left(\cfrac ne \right)^n$

where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.

Solving for strength coefficient

From the strain hardening equation we can solve for K

$K = \cfrac{S_{ut}}{\left(\cfrac ne \right)^n}$

And we can replace values

$K = \cfrac{275}{\left(\cfrac {0.4}e \right)^{0.4}}\\K=591.87$

Thus we get that the strength coefficient is K = 591.87 MPa

6 0

3 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$