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3 years ago

7

Damien wants to study the effect on materials let 60 after it is submerged in water for varying durations what type of graphical

tool should he use

1 answer:

Crank3 years ago

7 0

Answer:

please help you are not the intended recipient

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Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and

lina2011 [118]

Answer:

a.

y[n] = x[n] x[n-1] x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is $\frac{1}{x[n] . x[n-1] x[n+1]}$

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

for x[n] = 2 $\pi$ , y[n] = cos(2 $\pi$ ) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

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3 years ago

The loneliest people are to kindest

valkas [14]

Answer:

The most damaged people are the wisest is a fact

Explanation:

5 0

3 years ago

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A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o

victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width $= 25 mm = 25\times 10^{-3} m$

thickness $= 6.5 mm = 6.5\times 10^{-3} m$

crack length 2c = 0.5 mm at centre of specimen

$\sigma _{applied} = 1000 N/cross sectional area$

stress intensity factor = k will be

$\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}$

$= 6.154\times 10^{6} Pa$

we know that

$k =\sigma_{applied} (\sqrt{\pi C})$

$=6.154\sqrt{\pi (2.5\times 10^{-04})}$ [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if $K_C = 1.15 Mpa m^{1/2}$ then load will be

$Kc = \sigma _{frac}(\sqrt{\pi C})$

$1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}$

$\sigma _{frac} = 41.04 MPa$

$load = \sigma _{frac}\times Area$

$load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N$

LAOD = 6669.86 N

3 0

3 years ago

A sewage lagoon that has a surface area of 10 ha and a depth of 1 m is receiving 8,640 m^3 /d of sewage containing 100 mg/L of b

Marysya12 [62]

Answer: Coefficient= 0.35 per day

Explanation:

To find the bio degradation reaction rate coefficient, we have

k= $\frac{(Cin)(Qin)-(Cout)(Qout)}{(Clagoon)V}$

Here, the C lagoon= 20 mg/L

Q in= Q out= 8640 m³/d

C in= 100 mg/L

C out= 20 mg/L

V= 10 ha* 1* 10

V= 10⁵ m³

So, k= $\frac{8640*100-8640*20}{20*10^5}$

k= 0.35 per day

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3 years ago

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givi [52]

Answer:

Kawasaki Ninja H2R – top speed: 222 mph. This one is another beast in the form of a bike. ...

MTT Turbine Superbike Y2K – top speed: 227 mph. This bike is one of the most powerful production motorcycles. ...

Suzuki Hayabusa – top speed: 248 mph. 1340cc

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3 years ago

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