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emmasim [6.3K]
3 years ago
7

PLZZZZZ HELP

Engineering
2 answers:
Makovka662 [10]3 years ago
6 0

Answer:

As you might imagine, people working in robotics have strong mathematics, science, programming, and systems analysis skills. But design also plays a big role. Being able to iterate, isolate problems, and prototype until you have just the right design is essential.

Explanation:

hope that helps

antoniya [11.8K]3 years ago
5 0
I like to flush my girl friends tampons down the garbage disposal
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the voltage across a 5mH inductor is 5[1-exp(-0.5t)]V. Calculate the current through the inductor and the energy stored in the i
Serggg [28]

Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

V(t) = 5(1-e^{-0.5t})

The current flowing through the inductor is given by

i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\

i(t) = 1000t +2000e^{-0.5t} -2000\\

So the current at t = 2 seconds is

i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A

The energy stored in the inductor at t = 2 seconds is

W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J

4 0
3 years ago
Consider the cascade of the three LTI systems having impulse responses: h-1(t) = e^-tu(t + 3) h_2(t) = rect((1 -1)/2) h_3(t) = d
omeli [17]

Explanation:

There are two ways to find out the equivalent impulse response of the system.

1. Convolution in time domain

2. Simple multiplication in Laplace domain

2nd method is efficient, easy and is less time consuming.

Step 1: Take the Laplace transform of the given three impulse response functions to convert time domain signals into s-domain

Step 2: Once we get signals in s-domain, multiply them algebraically to get the equivalent s-domain response.

Step 3: Take inverse Laplace transform of the equivalent impulse response to convert from s-domain into time domain.

Solution using Matlab:

Step 1: Take Laplace Transform

Ys1 =  1/(s + 1)

Ys2 =  1/s - exp(-s/2)/s

Ys3 =  exp(-3*s)

Step 2: Multiplication in s-domain

Y =  (exp(-(7*s)/2)*(exp(s/2) - 1))/(s*(s + 1))

Step 3: Inverse Laplace Transform (Final Solution in Time Domain)

h =  heaviside(t - 7/2)*(exp(7/2 - t) - 1) - heaviside(t - 3)*(exp(3 - t) - 1)

8 0
4 years ago
A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac
Alex Ar [27]

Answer:

Throat diameter d_2=28.60 mm

Explanation:

 Bore diameter d_1=63mm  ⇒A_1=3.09\times 10^{-3} m^2

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004\frac{m^3}{s}

Coefficient of discharge C_d=0.8

We know that discharge through venturi meter

 Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

A_2=6.42\times 10^{-4} m^2

 ⇒d_2=28.60 mm

So throat diameter d_2=28.60 mm

     

4 0
3 years ago
Kaya just bought a house and realizes that the chimney needs to be totally torn down and rebuilt. She needs to call an expert si
lisabon 2012 [21]

a bricklayer, a bricklayer builds houses repairs walls and chimneys etc.

3 0
2 years ago
For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the
Goshia [24]

Answer:

\Delta p_{2} = 2\cdot \Delta p_{1}

Explanation:

The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.

\Delta p_{2} = 2\cdot \Delta p_{1}

8 0
3 years ago
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