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schepotkina [342]
3 years ago
14

Education is the foundation of profession. Justify​

Engineering
1 answer:
ololo11 [35]3 years ago
8 0

Answer:

Explanation:

Foundations of Education refers to a broadly-conceived field of educational and profoundly important academic and professional purpose unifies persons  Education as a profession in remarks made in relation to the expertise has long been the foundation of free and open academic inquiry.Setting goals for schools and students is an important process. It helps to motivate people when there is something specific to aim for. It is also a good way to show the clear path to success. The principal should work with the teachers to develop a plan and mission for the school

plz mark as brainliest

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he following is true for a Function Generator (select all that apply): Select one or more: a. It produces a variety of patterns
Amanda [17]

Answer:

A,C and E

Explanation:

It should be understood that a function generator can be explained as a piece of electronic test equipment or software which is used in generating different types of electrical waveforms over a wide range of frequencies. It should be noted that some of the examples of waveforms produced by the function generator are the sine wave, square wave, triangular wave and sawtooth shapes.

Therefore, the reason why of picking the options as highlighted above.

4 0
4 years ago
A teenage brain is already fully developed to enable us to manage risks effectively.
Kisachek [45]

Answer:

false

Explanation:

the brain is only really fully devolved by age 26

3 0
3 years ago
Read 2 more answers
A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.
Nataliya [291]

Answer:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

Explanation:

We can assume that the general formula for the drag force is given by:

D= C_D \frac{\rho}{2}V^2 A

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D1} \frac{\rho}{2}V^2 (wh) V

We can assume the same drag coeeficient C_{D1}=C_{D2}=C_{D} and we got:

\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D} \frac{\rho}{2}V^2 (wh) V

\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

\frac{l}{h}=\frac{5}{1.7}2.941

And we can estimate the calue of C_D = 1.2 from a figure.

And we can calculate the power difference like this:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

8 0
3 years ago
A metal shear can be used to cut flat stock , round stock , channel iron and which of the following?
Charra [1.4K]

Answer:

Angle Iron

Explanation:

Mark brainliest please.

3 0
3 years ago
Your boss asks you to classify some of the components in a temperature measurement system. The system consists of a thermocouple
lianna [129]

Answer:

Explanation:

1) Resolution and uncertainty of both ADC ranges

a) Resolution (for +/- 5 mV range) = 5 mV / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 mV / 256 = 0.0195 mV    ..................1

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*0.0195 mV = 0.00975 mV       ..........2

b) Resolution (for +/- 5 V range) = 5 V / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 V / 256 = 0.0195 V = 19.5 mV    ..................3

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*19.5 mV = 9.75 mV       ..........4

2)Thermocouple sensitivity = ( Maximum output voltage - Minimum Output voltage) / (Maximum Temperature - Minimum Temperature)

Thermocouple sensitivity = (3.649mv - 0 ) / (70 - 0) = 0.0521 mV / Deg.C           ............5

This is the required Thermocouple sensitivity

3) Water bath temperature is given as 57 deg.C

Hence voltage read by Thermocouple = Sensitivity*57 = 0.0521*57 mV = 2.9697 mV    ........6

4)We need to use ADC with a range of +/- 5 mV range as ADC with +/- 5 V range can not do measurement as it's resolution is higher than output voltage.

ADC will measure voltage as 2.9695 mV                    ......................7

8 0
3 years ago
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