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miv72 [106K]
4 years ago
11

a. A rectangular pen is built with one side against a barn. 1900 m of fencing are used for the other three sides of the pen. Wha

t dimensions maximize the area of the​ pen? b. A rancher plans to make four identical and adjacent rectangular pens against a​ barn, each with an area of 100 msquared. What are the dimensions of each pen that minimize the amount of fence that must be​ used?
Physics
1 answer:
Natalka [10]4 years ago
6 0

Answer:

475 m , 950 m

Explanation:

Let l be the length of the side perpendicular to the barn.

1900-2l = length of the side parallel to the barn

Area A= l( 1900-2l)

A= 1900l-2l^2

now, the maximum value of l ( the equation being quadratic)

l_max= -b/2a

a= 2

b=1900

l_max= -1900/4= 475 m

then 1900-2l= 1900-2×(475) = 950 m

So, the dimensions that maximize area are

950 and 475

Now. A_max = -2( l_max)^2+1900×l_max

A_max=  -2(475)^2+1900×475

A_max= 451250 m^2

or, 475×950 = 451250 m^2

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NeTakaya

Answer:

5.2791264*10¹³

Explanation:

Convert the 9 years to seconds and then multiple it by 186000

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3 years ago
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Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be
avanturin [10]

Answer:

The mass rate of the cooling water required is: 1'072988.5\frac{kg}{h}

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

{m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}

Where w refers to the cooling water and s to the steam flow. Reorganizing,

m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}

Write the difference of enthalpy for water as Cp (Tout-Tin):

m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, {h_{s}}^{out}=251.40\frac{kJ}{kg} and {h_{s}}^{in} can be calculated as:

{h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}.

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 \frac{kJ}{kgK}. If the average temperature is actually different, it won't mean a considerable mistake. Also we know that {T_{w}}^{out}-{T_{w}}^{in}\leq 10, so let's work with the limit case, which is {T_{w}}^{out}-{T_{w}}^{in}=10 to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}

Download pdf
5 0
3 years ago
A 4.00 kg ball is swung in a circle on the edge of a 1.50 m rope. The time it takes for the ball to complete one rotation is 3.4
lorasvet [3.4K]

Answer:

The answer is below

Explanation:

The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.

a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:

Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters

The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77  m/s

b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):

v = u + at

2.77 = 3.4a

a = 0.82 m/s²

c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N

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3 years ago
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Rainbow [258]

Answer:

The force is 15N

Explanation:

The formula is Force= mass × velocity.

From the question mass is 5kg, velocity is 3m/s.

F= 5×3

F= 15Newton.

Therefore the force is 15N.

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3 years ago
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