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4 years ago

Which type of engineering design uses an already-existing design?

Engineering

2 answers:

Vaselesa [24]4 years ago

8 0

Answer:

Explanation:

Arte-miy333 [17]4 years ago

4 0

The answer is D I’m 90% sure

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Water is leaking out of an inverted conical tank at a rate of 11700.0 cubic centimeters per min at the same time that water is b

Anastaziya [24]

Answer:

$V'_{in}=136,922.92cm^{3}/min$

Explanation:

In order to solve this problem, we must start by drawing the situation so we can see what the problem is about. (See attached picture)

So, since the problem wants us to write our answer in centimeters per minute, we can start by converting each of the meters to centimeters,so we get:

$5m*\frac{100cm}{1m}=500cm$

$14m*\frac{100cm}{1m}=1400cm$

$2.5m*\frac{100cm}{1m}=250cm$

Next, the volume of a cone can be found by using the following formula:

$V=\frac{1}{3} \pi r^{2}h$

In this case, both the radius and the height will be constantly changing, so we will need to find an equation to relate the height with the radius. We can find it by analyzing the cross section of the cone, which makes a right triangle. By using similar triangles we get:

$\frac{h}{r}=\frac{1400}{250}$

when solving for r we get that:

$r=\frac{250}{1400} h$

which simplifies to:

$r=\frac{5}{28} h$

so now, we can substitute this on our volume equation so we get:

$V=\frac{1}{3} \pi (\frac{5}{28}h)^{2}h$

which simplifies to:

$V=\frac{25}{2352}\pi h^{3}$

now we can take the derivative of that equation so we get:

$\frac{dV}{dt}=\frac{25}{784}\pi h^{2} \frac{dh}{dt}$

and now we can substitute:

$\frac{dV}{dt}=\frac{25}{784}\pi (250cm)^{2} (20cm/min)$

which solves to

$\frac{dV}{dt}=125,222.92\frac{cm^{3}}{min}$

this will represent the total amount of water that is being kept inside the tank. Now we can calculate the amount of water that is entering the tank.

$V'_{in}-V'_{out}=V'_{tot}$

so:

$V'_{in}=V'_{tot}+V'_{out}$

$V'_{in}=125,222.92cm^{3}/min+11,700cm^{3}/min$

$V'_{in}=136,922.92cm^{3}/min$

7 0

3 years ago

Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E

White raven [17]

Answer:

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

Explanation:

To solve this problem we use the expression for the temperature film

$T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5$

Then, we have to compute the Reynolds number

$Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}$

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

$Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77$

but we also now that

$Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\$

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

$Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J$

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0

4 years ago

A ball A is thrown vertically upward from the top of a 24-m-high building with an initial velocity of 7 m/s . At the same instan

solniwko [45]

Answer:

The height from ground at which they pass each other is 22.656 m

The time at which they pass each other is 1.6 sec.

Explanation:

For ball A, we have:

height = (h)a

Initial Velocity = Vi = 7 m/s

g = - 9.8 m/s² (for upward motion)

Using 2nd eqn. of motion:

h = Vi t + (1/2)gt²

(h)a = 7t + (1/2)(-9.8)t²

(h)a = 7t - 4.9t²

this is the height of ball A with reference as the building. Taking ground as reference, we have to add the height of building, that is, 24 m.

(h)a = 7t - 4.9t² + 24 ______ eqn (1)

For ball B, we have:

height = (h)b

Initial Velocity = Vi = 22 m/s

g = - 9.8 m/s² (for upward motion)

Using 2nd eqn. of motion:

h = Vi t + (1/2)gt²

(h)b = 22t + (1/2)(-9.8)t²

(h)b = 22t - 4.9t² ______ eqn (2)

Now, when the too balls pass each other, there height must be same.

Therefore,

(h)a = (h)b

using eqn (1) and eqn (2):

7t - 4.9t² + 24 = 22t - 4.9t²

22t - 7t = 24

t = 24/15

t = 1.6 sec

Now, for the height, at which they pass each other put t = 6sec in eqn (2)

Therefore,

h = (22)(1.6) - (4.9)(1.6)²

h = 35.2 - 12.544

h = 22.656 m

6 0

4 years ago

The ampere draw of a 5000 watt electric heater used on 120 volts is

Salsk061 [2.6K]

Answer:

41.67amps

Explanation:

v*a=w

120*a=5000

5000/12=41.67

8 0

3 years ago

Wind blows on the side of a fully enclosed hospital located on open flat terrain where V= 120 mi/h. Determine the external press

ss7ja [257]

Answer:

The external pressure is p = -21.9 psf or p = -8.85 psf

Explanation:

Given :

Velocity of wind, v = 120 mi / hr

$k_d = k_c =1$ (wind direction factor)

$k _{zt} = 1$ = topographical factor (for flat terrain)

$q_n$ = velocity pressure at height h

$\therefore q_n = 0.00256 k_z k_{zt} k_d v^2$

$q_n = 0.00256 \times k_z (1)(1)(120)^2$

$= 36.86 k_z$

But for height h = 30 ft, $k_z$ = 0.98 (from table)

$\therefore q_n = 36.86 \times 0.98$

= 36.16

Now, $\frac{L}{B}= \frac{200}{200} =1$ , so $C_p=-0.5$ (from table)

$p = q(G)(C_p)-q_n(GC_{pi})$

where, p = external pressure

G = 0.85 = gust factor (for typical rigid building)

$GC_{pi} = \pm 0.18$ (internal pressure co efficient)

Therefore putting the values,

$p = (36.13)(0.85)(-0.5)-(36.13)(\pm 0.18)$

p = -21.9 psf or p = -8.85 psf

4 0

4 years ago