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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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La iluminación de la superficie de un patio amplio es 1600 lx cuando el ángulo de elevación del sol 53°. Calcular la iluminación
gregori [183]

Answer:

 I = 1205.69 Lx

Explanation:

The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀

          I = I₀ sin θ

in this case with the initial data we can calculate the initial irradiance

         I₀ = \frac{I}{sin  \ \theta }

         I₀ = 1600 /sin 53

         I₀ = 2003.42 lx

for when the angle is θ = 37º

         I = 2003.42 sin 37

         I = 1205.69 Lx

6 0
3 years ago
A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is
jonny [76]

Answer:

a) t = 277.477\,s\,(4.625\min), b) v_{f} = 0\,\frac{mi}{h}, c) a = -0.128\,\frac{ft}{s^{2}}

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}

a = -0.185\,\frac{ft}{s^{2}}

The time required to travel is:

t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }

t = 277.477\,s\,(4.625\min)

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be v_{f} = 0\,\frac{mi}{h}.

c) The final constant deceleration is:

a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}

a = -0.128\,\frac{ft}{s^{2}}

7 0
3 years ago
Write the output expression for a NOR gate with inputs , , and .​
mezya [45]

Answer:

Logic NOR Gate Equivalent

The Boolean expression for a logic NOR gate is denoted by a plus sign, ( + ) with a line or Overline, ( ‾‾ ) over the expression to signify the NOT or logical negation of the NOR gate giving us the Boolean expression of: A+B = Q.

4 0
2 years ago
What is the maximum thermal efficiency possible for a power cycle operating between 600P'c and 110°C? a). 47% b). 56% c). 63% d)
Anastasy [175]

Answer:

(b) 56%

Explanation:

the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature

here we have given T₁ (Higher temperature)= 600+273=873

lower temperature T₂=110+273=383

Efficiency of power cycle is given by =1-\frac{T2}{T1}

=1-\frac{383}{873}

=1-0.43871

=.56

=56%

5 0
3 years ago
Current density is given in cylindrical coordinates as J = −106z1.5az A/m2 in the region 0 ≤ rho ≤ 20 µm; for rho ≥ 20 µm, J = 0
Naily [24]

Question:

Current density is given in cylindrical coordinates as J = −10^6z^1.5az A/m² in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0.

(a) Find the total current crossing the surface z = 0.1 m in the az direction.

(b) If the charge velocity is 2 × 10^6 m/s at z = 0.1 m, find ρν there.

(c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.

Answer:

a. -39.8μA

b. -15.81mC/m³

c. 29.05m/s

Explanation:

Given

Density = J = −10^6z^1.5az A/m²

Region: 0 ≤ ρ ≤ 20 µm

ρ ≥ 20 µm

J = 0.

a. Total current is calculated by.

J * ½((ρ1)² - (ρ0)²) * 2 π * φdza.

Where J = Density = -10^6 * z^1.5

ρ1 = Upper bound of ρ = 20

ρ0 = Lower bound of ρ = 0

π = 22/7

φdza = 10^-6

z = 0.1

Total current

= -10^6 * z^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= 10^6 * 0.1^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= −39.7543477278310

= -39.8μA

b. Calculating velocity charge density at (ρv)

Density (J) = ρv * V

Where J = Density = -10^6 * z^1.5

V = 2 * 10^6

z = 0.1

Substitute the above values

-10^6 * 0.1 ^1.5 = ρv * 2 * 10^6

ρv = (-10^6 * 0.1^1.5)/(2 * 10^6)

ρv = -0.1^1.5/(2)

ρv = -0.015811388300841

ρv = -0.01581 --------- Approximated

ρv = -15.81mC/m³

c. Calculating Velocity

Velocity = J/V

Where Velocity Charge Density = -2000 C/m3

Where J = -10^6 * z^1.5

z = 0.15

J = -10^6 * 0.15^1.5

J = -58094.75019311125

Velocity = -58094.75019311125/-2000

Velocity = 29.047375096555625m/s

Velocity = 29.05m/s

8 0
3 years ago
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