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3 years ago

Hey, can anyone tell me if Igneous rock is good to build on? Cheers!

Engineering

1 answer:

lorasvet [3.4K]3 years ago

6 0

Answer:

Yes! Here's Why

Explanation:

Igneous rocks are often used as setts in areas of heavy wear, because they withstand both weathering and erosion better than most sedimentary rocks, or even concrete.

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Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a

nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

6 0

4 years ago

Ok there........................................................................

Juliette [100K]

Answer:

ok THERE

Explanation:

4 0

3 years ago

Tech A says that multiple gasket or seal leaks can be caused by a defective exhaust gas recirculation (EGR) system.

lesya [120]

I think the answer to this question is b but I’m not sure

7 0

3 years ago

A disk of radius 2.1 cm has a surface charge density of 5.6 µC/m2 on its upper face. What is the magnitude of the electric field

Assoli18 [71]

Answer:

$=6.3*10^3 N/C$

Explanation:

solution:

from this below equation (1)

$E=$ σ/2εo $(1-\frac{z}{\sqrt{z^2-R^2} } )$ ...........(1)

we obtain:

$=5.6*10^-6 \frac{c}{m^2} /2(8.85*10^-12\frac{c^2}{N.m^2} ).(1-\frac{9.5 cm}{\sqrt{9.5^2-2.1^2} } )$

$=6.3*10^3 N/C$

8 0

3 years ago

Given an integer k, a set C of n cities c1, . . . , cn, and the distances between these cities dij = d(ci , cj ), for 1 â¤ i &lt

Snezhnost [94]

Answer:

See explaination

Explanation:

2-Approximation Algorithm

Step 1: Choose any one city from the given set of cities C arbitrarily and put it in to a set H which is initially empty.

Step 2: For every city c in set C that is currently not present in set H compute min_distc = Minimum[ d(c, c1), d(c, c2), d(c, c3), ..... . . . . d(c, ci) ]

where c1, c2, ... ci are the cities in set H

and d(x, y) is the euclidean distance between city x and city y

Step 3: H = H ∪ {cx} where cx is the city have maximum value of min_dist over all possible cities c, computed in Step-2.

Step 4: Step-2 and Step-3 are iterated for k-1 times so that k cities are included int set H.

The set H is the required set of cities.

Example

Assume:-

C = {0, 1, 2, 3}

d(0,1) = 10, d(0,2) = 7, d(0,3) = 6, d(1,2) = 8, d(1,3) = 5, d(2,3) = 12

k = 3

Solution:-

Initially H = { }

Step-1: H = {0}

Step-2: Cities c \not\in H are {1, 2, 3}

min_dist1 = min{dist(0,1)} = min{10} = 10

min_dist2 = min{dist(0,2)} = min{7} = 7

min_dist3 = min{dist(0,3)} = min{6} = 6

Step-3: Max{10, 7, 6} = 10

Step-4: cx = 1

Step-5: H = H ∪ cx = {0} \cup {1} = {0, 1}

Step-6: Cities c \not\in H are {2, 3}

min_dist2 = min{dist(0,2), dist(1,2)} = min{7, 8} = 7

min_dist3 = min{dist(0,3), dist(1,3)} = min{6, 5} = 5

Step-7: Max{7, 5} = 7

Step-8: cx = 2

Step-9: H = H \cup cx = {0, 1} \cup {2} = {0, 1, 2}

Result: The set H is {0, 1, 2}.

6 0

3 years ago