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2 years ago

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Assume (for simplicity in this exercise) that only one tuple fits in a block and memory holds at most three blocks. Show the run

s created on each pass of the sort-merge algorithm when applied to sort the following tuples on the first attribute: (kangaroo, 17), (wallaby, 21), (emu, 1), (wombat, 13), (platypus, 3), (lion, 8), (warthog, 4), (zebra, 11), (meerkat, 6), (hyena, 9), (hornbill, 2), (baboon, 12).

1 answer:

den301095 [7]2 years ago

6 0

Answer:

See explaination

Explanation:

Let's define tuple as an immutable list of Python objects which means it can not be changed in any way once it has been created.

Take a look at the attached file for a further detailed and step by step solution of the given problem.

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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

Timken rates its bearings for 3000 hours at 500 rev/min. Determine the catalog rating for a ball bearing running for 10000 hours

svet-max [94.6K]

Answer:

C₁₀ = 6.3 KN

Explanation:

The catalog rating of a bearing can be found by using the following formula:

C₁₀ = F [Ln/L₀n₀]^1/3

where,

C₁₀ = Catalog Rating = ?

F = Design Load = 2.75 KN

L = Design Life = 1800 rev/min

n = No. of Hours Desired = 10000 h

L₀ = Rating Life = 500 rev/min

n₀ = No. of Hours Rated = 3000 h

Therefore,

C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3

C₁₀ = (2.75 KN)(2.289)

<u>C₁₀ = 6.3 KN</u>

3 0

3 years ago

Why is it that dislocations play an important role in controlling the mechanical properties of metallic materials, however, they

klio [65]

Answer:

dislocations play an important role in controlling as

Explanation:

As dislocations plays an important role in the ductility, elasticity and plurality of materials

The elastic and elastic deflections play a large role in their properties as the metallic materials, because the dislocation of a glass material does not play a major role in their properties.

5 0

2 years ago

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o

jekas [21]

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu$

$\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

$H_m=(\frac{200*10^3}{1000*9.8}-10)$

$H_m=10.39m$

Generally the equation for Power is mathematically given by

$P=\rho gQH_m$

Therefore

$Q=\frac{P}{\rho g H_m}$

$Q=\frac{4*10^4}{1000*9.81*10.9}$

$Q=0.03935m^3/sec$

Since

$Q=AV'$

Where

$A=\pi r^2\\A=3.142 (0.05)^2$

$A=7.85*10^{-3}$

Therefore

$V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

$V'=5m/s$

5 0

3 years ago

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

3 years ago