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Fudgin [204]
3 years ago
12

Câu 1: Tìm từ trong 2 câu thơ sau liên quan đến các từ: hình chiếu, tia chiếu, mặt phẳng chiếu.

Engineering
1 answer:
xeze [42]3 years ago
4 0

Explanation:

are bhai brainly aap english me questions kar ne ko hai

ye kon si bhasha hai ??????

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Knowing that v = –8 m/s when t = 0 and v = 8 m/s when t = 2 s, determine the constant k. (Round the final answer to the nearest
docker41 [41]

Answer:

a)We know that acceleration a=dv/dt

So dv/dt=kt^2

dv=kt^2dt

Integrating we get

v(t)=kt^3/3+C

Puttin t=0

-8=C

Putting t=2

8=8k/3-8

k=48/8

k=6

5 0
3 years ago
Consider the following ways of handling deadlock: (1) banker’s algorithm, (2) detect
Andrew [12]

Answer:

b

Explanation:

7 0
3 years ago
Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a c
vlabodo [156]

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

COP_{i} = \frac{250\,K}{300\,K-250\,K}

COP_{i} = 5

The real Coefficient of Performance is:

COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}

COP_{r} = 12.571

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

6 0
3 years ago
Best use for a suspension bridge
babunello [35]

Answer:

over a rive or fast moving water or canyon

Explanation: you would use a suspension bridge in an area where you can't put supports down.

4 0
3 years ago
Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average valu
Varvara68 [4.7K]

The question is not complete. We are supposed to find the average value of v_o.

Answer:

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

3/(T/4) = 0.7/t1

So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)

v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)

v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

8 0
3 years ago
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