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2 years ago

How do I get my son to do his work?

Engineering

1 answer:

valentinak56 [21]2 years ago

8 0

Answer:

Explanation:

Reward him if he does his homework/work

If he doesnt do his homework, take things that he loves off of him. and tell him if he does his homework/work he will get them things back

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Cho biết tác dụng chung của các hệ giằng khung ngang nhà công nghiệp nhẹ 1 tầng 1 nhịp.

Dmitry_Shevchenko [17]

I don’t know how to speak the laungue or know this language

4 0

2 years ago

What is the maximal coefficient of performance of a refrigerator which cools down 10 kg of water (and then ice) to -6C. Upper he

inysia [295]

Given:

Temperature of water, $T_{1}$ = $-6^{\circ}C$ =273 +(-6) =267 K

Temperature surrounding refrigerator, $T_{2}$ = $21^{\circ}C$ =273 + 21 =294 K

Specific heat given for water, $C_{w}$ = 4.19 KJ/kg/K

Specific heat given for ice, $C_{ice}$ = 2.1 KJ/kg/K

Latent heat of fusion, $L_{fusion}$ = 335KJ/kg

Solution:

Coefficient of Performance (COP) for refrigerator is given by:

Max $COP_{refrigerator}$ = $\frac{T_{2}}{T_{2} - T_{1}}$

= $\frac{267}{294 - 267}$ = 9.89

Coefficient of Performance (COP) for heat pump is given by:

Max $COP_{heat pump}$ = $\frac{T_{1}}{T_{2} - T_{1}}$ $\frac{294}{294 - 267}$ = 10.89

6 0

3 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

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during

astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

$\sigma_c$ = tensile stress = 200 Mpa

$K$ = plane strain fracture toughness= 55 Mpa $\sqrt{m}$

$\lambda$ = length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

$\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}$ (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for $\lambda$

Multiplying both sides of equation (1) by $Y\sqrt{\pi\lambda}$

$\sigma_c Y\sqrt{\pi\lambda}=K$ (2)

Sequaring both sides of equation (2):

$(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2$

$\sigma^2_c Y^2 \pi\lambda=K^2$ (3)

Dividing both sides by $\sigma^2_c Y^2 \pi$ we got:

$\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2$ (4)

Replacing the values into equation (4) we got:

$\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m$

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

7 0

3 years ago

When you learned about the Highway Transportation system you learned about 6 different types of people as

Musya8 [376]

Answer: 3 different types of people using Highway Transportation system are :

1. Pedestrians - should be given priority while driving.

2. Cyclists - should be given enough space on road.

3. Motorcyclists- should be given enough space on road.

Explanation:

1. Pedestrians - Most of pedestrians use to walk on the footpath along side road that keeps them on a safe side. But there are places without footpath along side road, in that case they have to walk on the road itself. Here, we need to take care for them. We need to wait in case they are crossing road and also check for them while taking a turn.

2. Cyclists - They travel on road but are tough to figure out. They travel at a slower pace compared to cars. To avoid any accident with them, we are supposed to give them enough space which should be equivalent to a car's space.

3. Motorcyclists - They can pass by very closely and also come between lanes. Most of the things to be considered here are same as that of cyclists. Here also, we need to check for them carefully while taking a turn. Also, need to give them enough space.

4 0

2 years ago