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Sophie [7]
3 years ago
7

Two ways of sawing logs are?​

Engineering
1 answer:
zmey [24]3 years ago
7 0

Answer:

Plain Sawn. At least 95 percent of all hardwood lumber commercially produced in the U.S. is flat or plain sawn. ...

Quarter Sawn. In this method, the log actually is cut into quarters, then sawn quarter by quarter. ...

Explanation:

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A polymeric extruder is turned on and immediately begins producing a product at a rate of 10 kg/min. An operator realizes 20 min
hodyreva [135]

Answer:

The plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.

The production rate function M(t) is:

m(t)=[H(t)\cdot10+H(t-20)\cdot5-H(t-80)\cdot14+H(t-81)\cdot9]kg/min (1)

The Laplace transform of this function is:

\displaystyle m(s)=[\frac{10+5e^{-20s}-14e^{-80s}+9e^{-81s}}{s}]kg/min    (2)

Explanation:

The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.

For the Laplace transform we use the following rules:

\mathcal{L}[f(x)+g(x)]=\mathcal{L}[f(x)]+\mathcal{L}[g(x)]=F(s)+G(s)    (3)

\mathcal{L}[aH(x-b)]=\displaystyle\frac{ae^{-bs}}{s}    (4)

4 0
4 years ago
A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a
statuscvo [17]

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

3 0
3 years ago
When you apply for your driver license, you consent to take a ____ test when asked to do so by a law enforcement officer. memory
n200080 [17]

Answer:

Driving test

Explanation:

Usually according to laws in countries worldwide, to be licenced to drive, one is required to go through a driving school to learn the ethics and rules of driving.

4 0
2 years ago
Consider the database table structure shown in the figure. (a) Write a SELECT statement statement (compatible with an Oracle RDB
Delicious77 [7]

Answer:

The table is attached as a picture.

a)

Select VENDOR_CONTACT_LAST_NAME || ', ' || VENDOR_CONTACT_FIRST_NAME "full_name" from VENDORS where VENDOR_CONTACT_LAST_NAME like 'A%' or VENDOR_CONTACT_LAST_NAME like 'E%' order by VENDOR_CONTACT_LAST_NAME,VENDOR_CONTACT_FIRST_NAME;

concatenation operator || is used . Also LIKE is used for pattern matching. full_name is alias for the concatenated column

b) As sample data is not given ,Please test the query for the data given in table

Explanation:

6 0
3 years ago
The output S/N at thereceiver must be greater than 40 dB. The audio signal has zero mean, maximum amplitude of 1, power of ½ Wan
abruzzese [7]

Given that,

The output signal at the receiver must be greater than 40 dB.

Maximum amplitude = 1

Bandwidth = 15 kHz

The power spectral density of white noise is

\dfrac{N}{2}=10^{-10}\ W/Hz

Power loss in channel= 50 dB

Suppose, Using DSB modulation

We need to calculate the power required

Using formula of power

P_{L}_{dB}=10\log(P_{L})

Put the value into the formula

50=10\log(P_{L})

P_{L}=10^{5}\ W

For DSB modulation,

Figure of merit = 1

We need to calculate the input signal

Using formula of FOM

FOM=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}

1=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}

\dfrac{S_{i}}{N_{i}W}=\dfrac{S_{o}}{N_{o}}

Put the value into the formula

\dfrac{S_{i}}{2\times10^{-10}\times15\times10^{3}}

\dfrac{S_{i}}{30\times10^{-7}}

S_{i}

S_{i}=30\times10^{-3}

We need to calculate the transmit power

Using formula of power transmit

S_{i}=\dfrac{P_{t}}{P_{L}}

P_{t}=S_{i}\times P_{L}

Put the value into the formula

P_{t}=30\times10^{-3}\times10^{5}

P_{t}=3\ kW

We need to calculate the needed bandwidth

Using formula of bandwidth for DSB modulation

bandwidth=2W

Put the value into the formula

bandwidth =2\times15

bandwidth = 30\ kHz

Hence, The transmit power is 3 kW.

The needed bandwidth is 30 kHz.

3 0
3 years ago
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