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4 years ago

Please help will give brainliest please answer all 3

Engineering

1 answer:

larisa [96]4 years ago

8 0

Answer: For #1 I'm going to go with A because that has to do with biology

For #2 I'm going to go with B oceans because that has to do with plant life (and life in general).

For #3 I'll say marine/maritime engineer (you can just say marine)

Hope it helps!

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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

1) Find the time in seconds to reach full charge in an RL circuit with L = 5 H and R = 100 ohms

pav-90 [236]

The time constant to reach full charge in an RL circuit is 0.05 ms.

Explanation:

To find the time constant,

The time constant for an RL circuit is defined by τ = L/R.

The given data is

L= 5 H

R= 100 ohms

by using the formula,

τ = L/R

= 5/100

= 0.05 ms

τ = 0.05 ms

Thus, the time constant to reach full charge in an RL circuit is 0.05 ms.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=2x%20-%2014x%20%3D%2010" id="TexFormula1" title="2x - 14x = 10" alt="2x - 14x = 10" align="abs

tester [92]

Ain’t this math why is on here

3 0

3 years ago

Read 2 more answers

13–27. The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-k

Feliz [49]

Answer:

See explanation for step by step procedure to get answer.

Explanation:

Given that:

The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.

See the attachments for complete steps to get answer.

4 0

4 years ago

A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u

Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

$\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}$

where,

$\gamma$ = Specific Heat ratio of air

M = Mach number upstream

We know that $\gamma _{air}=1.4$

Applying values we get

$\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)$

Similarly the temperature downstream is obtained by the relation

$\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}$

Applying values we get

$\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C$

The Mach number downstream is obtained by the relation

$M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616$

3 0

4 years ago