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mrs_skeptik [129]
3 years ago
13

Coving is a curved edge between a floor and a wall. O True O False

Engineering
2 answers:
wolverine [178]3 years ago
7 0

Answer:

True.

Explanation:

ser-zykov [4K]3 years ago
3 0
Uhmmmmmmm it’s True..
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Sharon is designing a house in an area that receives a lot of rainfall all year. Which material should she use to stick the wood
kakasveta [241]

Explanation:

She is passionate about architecture, typography, and black & white film ... Since moving to Texas, I've heard a lot of people say, "If you don't like ... Oc, 3.74, 56, 80 ... Not only does the weather have to be clear to pour the concrete, but it ... system that goes within the slab) is complete, any additional rain will

4 0
3 years ago
The greater the force applied to an object, the _____ the change in speed or direction of the object.
storchak [24]

Answer:

b

Explanation:

8 0
2 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

5 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
max2010maxim [7]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

7 0
2 years ago
A rocket is launched from rest with a constant upwards acceleration of 18 m/s2. Determine its velocity after 25 seconds
lisabon 2012 [21]

Answer:

The final velocity of the rocket is 450 m/s.

Explanation:

Given;

initial velocity of the rocket, u = 0

constant upward acceleration of the rocket, a = 18 m/s²

time of motion of the rocket, t = 25 s

The final velocity of the rocket is calculated with the following kinematic equation;

v = u + at

where;

v is the final velocity of the rocket after 25 s

Substitute the given values in the equation above;

v = 0 + 18 x 25

v = 450 m/s

Therefore, the final velocity of the rocket is 450 m/s.

5 0
2 years ago
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