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Julli [10]
2 years ago
7

Which of the following is not a relationship set between elements in a sketch​

Engineering
1 answer:
Svet_ta [14]2 years ago
6 0

Answer:

The entity relationship (ER) data model has existed for over 35 years. It is well suited to data modelling for use with databases because it is fairly abstract and is easy to discuss and explain. ER models are readily translated to relations. ER models, also called an ER schema, are represented by ER diagrams.

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To revise a monthly budget, changes in which categories might need to be addressed? Check all that apply.
GREYUIT [131]

Income

amount budgeted

expenses

5 0
2 years ago
The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be
lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = \frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}       .....................1

Q = \frac{T \infty - Tso}{\frac{1}{hA}}                         .....................2

now we compare both equation 1 and 2 and put here value

\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}            

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

3 0
2 years ago
Even though the content of many alcohol blends doesn’t affect engine drive ability using gasoline with alcohol in warm weather m
Alchen [17]
Even though the content of many alcohol blends doesn't affect engine driveability, using gasoline with alcohol in warm weather may cause: decrease in fuel economy.

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3 0
2 years ago
A car accelerates from O to 60. miles per hour in 5.2 seconds. Calculate acceleration in m/s seconds. Calculate acceleration in
DochEvi [55]

Answer:

a=5.515\frac{m}{s^{2} }

Explanation:

The first thing we will do is convert the units. Miles per hour to meters per second.

1 mile=1609.34 mts.

1 hora=3600 segundos

Performing the operations

60\frac{mile}{h}=\frac{(60*1609.34)}{3600}\frac{m}{s}=26.822\frac{m}{s}

Now, we will use the acceleration formula

a=\frac{v}{t}

Where v = speed and t = time

Substituting the values ​​of t=5.2s

a=\frac{v}{t} =\frac{26.822\frac{m}{s} }{5.2s} =5.15\frac{m}{s^{2} }

7 0
3 years ago
. Two rods, with masses MA and MB having a coefficient of restitution, e, move
GarryVolchara [31]

Answer:

a) V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}

b) U_A = 3.66 m/s

V_B = 4.32 m/s

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let U_A be the initial velocity of rod A

Let U_B be the initial velocity of rod B

Let V_A be the final velocity of rod A

Let V_B be the final velocity of rod B

Using the principle of conservation of momentum:

M_AU_A + M_BU_B = M_AV_A + M_BV_B............(1)

Coefficient of restitution, e = \frac{V_B - V_A}{U_A - U_B}

V_A = V_B - e(U_A - U_B)........................(2)

Substitute equation (2) into equation (1)

M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B..............(3)

Solving for V_B in equation (3) above:

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}....................(4)

From equation (2):

V_B = V_A + e(U_A -U_B)......(5)

Substitute equation (5) into (1)

M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))..........(6)

Solving for V_A in equation (6) above:

V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}.........(7)

b)

M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?

Rod A is said to be at rest after the impact, V_A = 0 m/s

Substitute these parameters into equation (7)

0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s

To calculate the final velocity, V_B, substitute the given parameters into (4):

V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s

c) Impulse, I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)

I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))

I = 0 kg m/s^2

d) %\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%

%\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%

% \triangle KE = -47.85 \%

7 0
3 years ago
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