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Arisa [49]
3 years ago
14

An uncovered swimming pool loses 1.0 inch of water off its 1,000 ft^2 surface each week due to evaporation. The heat of vaporiza

tion for water at the pool temperature is 1050 btu/lb. The cost of energy to heat the pool is $10.00 per million btu. A salesman claims that a S500 pool cover that reduces evaporation losses by two-thirds will pay for itself in one 15-week swimming season. Can it be true?
Engineering
1 answer:
soldi70 [24.7K]3 years ago
3 0

Answer:

The affirmation is true, the cover will be worth buying

Explanation:

The equation necessary to use is

E = m*cv,

Where

cv: the heat of vaporization.  

Finding the rate at which the water evaporates (m^3/week).  

The swimming pool loses water at 1 inch/week off its 1,000 ft^2

Than,

1000 ft² * 1 in/wk * 1 ft/12 in = 83.33 ft³/week

To obtains the rate of mass loss it is necessary to multiply it for the density of water

83.33 ft³/week * 62.4 lb/ft³ = 5200 lb/week

Knowing the vaporization heat it is possible to find the rate of heat which is leaving the swimming pool  

5200 lb/week * 1050 BTU/lb = 5460000 btu/week

Over a 15-week period, the pool loses 81.9 million BTU.  

Knowing the cost of energy to heat the pool is $10.00 per million btu

The price = $819

This way, the affirmation is true, the cover will be worth buying

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The toggle (t) flip-flop has one input, clk, and one output, q. on each rising edge of clk, q toggles to the complement of its p
jeyben [28]

Answer:

  See attached

Explanation:

The next state of a toggle flip-flop is the inverse of the present state. This behavior can be produced using a D flip-flop that has its input connected to the inverse of its output.

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A schematic is attached.

7 0
2 years ago
The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
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alekssr [168]

Answer:

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Explanation:

8 0
3 years ago
Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1
tia_tia [17]

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is :  (n - 1) × L/R

where L :  packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate =  2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

4 0
3 years ago
In a short essay, discuss the question, "How are you an innovator?"
iragen [17]

Answer:

Being innovative means doing things differently or doing things that have never been done before. An innovator is someone who has embraced this idea and creates environments in which employees are given the tools and resources to challenge the status quo, push boundaries and achieve growth.

Explanation:

Hope it helps..

But it's a little bit long..

Correct me if I'm wrong..

7 0
3 years ago
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