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2 years ago

2. BCD uses 6 bits to represent a symbol. a) True b) False

Engineering

1 answer:

Goshia [24]2 years ago

4 0

Answer:

true because BCD used 6 bits to represent a symbol .

Explanation:

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Convert 250 lb·ft to N.m. Express your answer using three significant figures.

vfiekz [6]

Answer:

It will be equivalent to 338.95 N-m

Explanation:

We have to convert 250 lb-ft to N-m

We know that 1 lb = 4.45 N

So foe converting from lb to N we have to multiply with 4.45

So 250 lb = 250×4.45 =125 N

And we know that 1 feet = 0.3048 meter

Now we have to convert 250 lb-ft to N-m

So $250lb-ft=250\times 4.45N\times 0.348M=338.95N-m$

So 250 lb-ft = 338.95 N-m

6 0

3 years ago

The temperature of a system rises by 10 °C during a heating process. Express the rise in temperature of K, R, and °F.

Lorico [155]

Explanation:

Given T = 10 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (10 + 273.15) K = 283.15 K

T = 283.15 K

The conversion of T( °C) to T(F) is shown below:

T (°F) = (T (°C) × 9/5) + 32

So,

T (°F) = (10 × 9/5) + 32 = 50 °F

T = 50 °F

The conversion of T( °C) to T(R) is shown below:

T (R) = (T (°C) × 9/5) + 491.67

So,

T (R) = (10 × 9/5) + 491.67 = 509.67 R

T = 509.67 R

6 0

3 years ago

A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line throug

Elanso [62]

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

$\rho = \frac{m}{V}$

Where

m = Mass

V = Volume

For state one we know that

$\rho_1 = \frac{m_1}{V}$

$m_1 = \rho_1 V$

$m_1 = 1.18*1$

$m_1 = 1.18Kg$

For state two we have to

$\rho_2 = \frac{m_2}{V}$

$m_2 = \rho_2 V$

$m_1 = 7.2*1$

$m_1 = 7.2Kg$

Therefore the total change of mass would be

$\Delta m = m_2-m_1$

$\Delta m = 7.2-1.18$

$\Delta m = 6.02Kg$

Therefore the mass of air that has entered to the tank is 6.02Kg

5 0

3 years ago

It has a piece of 1045 steel with the following dimensions, length of 80 cm, width of 30 cm, and a height of 15 cm. In this piec

Serggg [28]

Answer:

material remove in 3 min is 16790.4 mm³/s

Explanation:

given data

length L = 80 cm = 800 mm

width W = 30 cm

height H = 15 cm

make grove length = 80 cm

width = 8 cm

depth = 10 cm

mill toll diameter = 4 mm

axial cutting depth = 20 mm

to find out

How much material removed in 3 minutes

solution

first we find time taken for length of advance that is

time = $\frac{length}{advance}$

here advance is given as 0.001166 mts / sec

so time = $\frac{800}}{0.001166*1000}$

time = 686.106 seconds

now we find material remove rate that is

remove rate = mill toll rate × axial cutting depth × advance

remove rate = 4 × 20×0.001166 ×1000

remove rate = 93.28 mm³/s

material remove in 3 minute = 3 × 60 = 180 sec

so material remove in 3 min = 180 × 93.28

material remove in 3 min is 16790.4 mm³/s

7 0

3 years ago

For a LED diode that has a= 632 nm, then the A1 is equal to:

alexgriva [62]

Answer:

1.693242

Explanation:

The colors in the Light emitting diodes have been identified by wavelength which is measured in nano-meters. Wavelength is a function of LED chip material. The LED diode which has a = 632 then A1 will be 1.63242, this is calculated by 1 / 632. Wavelength are important for human eye sensitivity. The colors emitted from the LED will depend on the semiconductor material.

5 0

2 years ago

2. BCD uses 6 bits to represent a symbol. a) True b) False​

2. BCD uses 6 bits to represent a symbol. a) True b) False