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2 years ago

2. In the above figure, what type of cylinder arrangement is shown in the figure above?

Engineering

2 answers:

sergeinik [125]2 years ago

7 0

The answer is c :) ..................................

ivann1987 [24]2 years ago

5 0

Answer: C. Horizontal

Explanation:
Remember engine layouts are usually characterized on their cylinder layout, valves and camshafts.

Types of engines:
•In-line engines have it’s cylinders arranged in a row (ex. In-line 4)

•V-Type engines have it’s cylinder banks are arranged at an angle to each other, so that the banks form a "V" shape when viewed from the front of the engine.

•Horizontal/flat engines have it’s cylinders located on either side of a central crankshaft. (ex. Boxer engine)

•L-Type engines, I’m going to be real I’ve never heard of a L-type layout and I don’t think it’s real.

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During genetic engineering, how do Restriction enzymes know what base pairs to act on?

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Answer:

The correct/closest option is b

Explanation:

Restriction enzymes are enzymes (endonucleases) that cut short DNA strands at specific sites. Hence, each restriction enzyme has it's own specific site (between two bases) it cuts at. There are two types of end that can be produced by this cut; the blunt end and the sticky end.

A restriction enzyme recognizes (palindromic sequence) and cut in it's own specific end.

For example, if a restriction enzyme cuts between a guanine (G) and an adenine (A), and it cuts a palindromic double stranded DNA in the manner below, it produces a sticky end.

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And if a restriction enzyme cuts between guanine (G) and cytosine (C) in the manner below, it produces a blunt end.

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2 years ago

How do you identify all sensors, functions, and where we can use them?

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2 years ago

The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k

sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

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