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Strike441 [17]
3 years ago
5

Find the sum and product of each of these pairs of numbers. Express your answers as a base 3 expansion. Hint: We can just add an

d multiply using the grade-school algorithms (working column by column starting at the right), using the addition and multiplication tables in base three (for example, 2 + 1 = 10 and 2 · 2 = 11). When a digit-by-digit answer is too large to fit (i.e., greater than 2), we "carry" into the next column. (10 points) (112)3, (210)3
Engineering
1 answer:
Arlecino [84]3 years ago
8 0

Answer:

112_3 + 210_3 = 1022_3

112_3 \times 210_3 = 101220_3

Explanation:

To find the sum of the two numbers, start summing from the right most numbers.

right most: 2 + 0 = 2

middle: 1 + 1 = 2

left: 1 + 2 = 3. Converting 3 to base 3, we have 1 r 0 = 10.

Therefore, 112_3 + 210_3 = 1022_3.

To find the product, simply do

(0 \times 112) + (1 \times 112) + (2 \times 112)\\ = 000 + 112 + 1001 = 101220.

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14. An engine is brought into the shop with a
Lostsunrise [7]

Answer:

B. To accurately measure spark advance, use a timing light that incorporates an

ignition advance meter. The spark advance cannot be determined by listening to the way the engine sounds.

8 0
2 years ago
What is the best way to submit your assignments?
Mrac [35]
A. Email your teacher right away. It would be the safest option.
4 0
3 years ago
Read 2 more answers
.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves
vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680m^3 and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 m^3/kg

Specific volume of saturated liquid=v_f=0.00102 m^3/kg

Now the relation  of total specific volume for a specific dryness value is given as

                                  v=v_f+x(v_g-v_f)

Substituting the values give

v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg

Now the volume of vessel is given as

v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3

So the volume of vessel is 4.7680m^3.

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680m^3 and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ

So the quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0
3 years ago
The drag force, Fd, imposed by the surrounding air on a
Ad libitum [116K]

Answer:

a)  23.551 hp

b)  516.89 hp

Explanation:

<u>given:</u>

F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2

<u>required:</u>

the power in hp

<u>solution:</u>

(F_{d})_{a}  =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}.............(1)

by substituting in the equation (1)

         =353.27 lbf

(F_{d})_{b}  =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}..........(2)

by substituting in the equation (2)

         = 2769.29 lbf

power is defined by

             P=F.V

     P_{a}=353.27*36.67

           =12954.411 lbf.ft/s

           =12954.411*.001818

           =23.551 hp

      P_{a}=2769.29*102.67

           = 284323 lbf.ft/s

           = 284323*.001818

           = 516.89 hp

3 0
3 years ago
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It is possible to generate a policy in which common points such as those mentioned above are agreed in order to hire or fire employees in their function of their psychological personality, that is, the character of knowledge and skills. Depending on the company, Test could be created in order to evaluate the psychological skills of the employees, as well as Test to periodically determine how their employees are kept up to date with regard to knowledge. The cumulative filter could be done every semester, for which each employee must exceed a minimum margin of score on these tests, otherwise his position could be at risk.

At the same time, incentives can be generated for the best scores that are rewarded not only with monetary values but also with rest days, coupons in restaurants or sports, which would cause the worker to strive to be constantly learning.

This policy agreement is outside the vision and mission of the company, and whose information must be given to the worker once he begins his work activities.

5 0
3 years ago
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