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3 years ago

B/ Evaluate e^(πi/2)

Engineering

2 answers:

ivanzaharov [21]3 years ago

8 0

Explanation:

≈4.8

There really isn't an elegant way to express it. Just plug and chug for irrationals raised to other irrationals.

Lilit [14]3 years ago

7 0

You get a result immediately from Euler's formula:

e ^(i π/2) = cos(π/2) + i sin(π/2) = 0 + i * 1 = i

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After a 65 newton weight has fallen freely from rest a vertical distance of 5.3 meters, the kinetic energy of the weight is

inysia [295]

Answer:

The kinetic energy of the weight is 344.5 J

Explanation:

Given that:

Force = F = 65 newton

distance = d = 5.3 meters

We have to find change in kinetic energy ΔK.E

Now we know that, initially kinetic energy was 0 So the formula we use will be:

Work done = Change in kinetic energy

Mathematically,

W = ΔK.E

As we know W = F . d and ΔK.E = K.E(final) - K.E(initial)

So by putting values:

F . d = K.E(final) - K.E(initial)

F . d = K.E(final)

As K.E(initial) is 0 so by putting values of F and d

(65)* (5.3) = K.E(final)

344.5 J = K.E(final)

So the change in K.E will also be 344.5 J

i hope it will help you!

3 0

3 years ago

I love touching the atmospheres crest

Lina20 [59]

That is super cool :))))

4 0

2 years ago

All of the following are examples of capital intensive industries EXCEPT: *

OverLord2011 [107]

i believe it is soft drink production

8 0

2 years ago

Read 2 more answers

A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l

Papessa [141]

Answer:

(a) 2.5 ksi

(b) 0.1075 in

Explanation:

(a)

$E=\frac {\sigma}{\epsilon}$

Making $\sigma$ the subject then

$\sigma=E\epsilon$

where $\sigma$ is the stress and $\epsilon$ is the strain

Since strain is given as 0.025% of the length then strain is $\frac {0.025}{100}=0.00025$

Now substituting E for $10\times 10^{6} psi$ then

$\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi$

(b)

Stress, $\sigma= \frac {F}{A}$ making A the subject then

$A=\frac {F}{\sigma}$

$A=\frac {\pi(d_o^{2}-d_i^{2})}{4}$

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that $2t=d_o - d_i$ where t is thickness

Now substituting

$\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}$

$\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4$

$(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

But the outer diameter is given as 2 in hence

$(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

$2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}$

$d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in$

As already mentioned, $2t=d_o - d_i hence t=0.5(d_o - d_i)$

$t=0.5(2-1.785)=0.1075 in$

3 0

4 years ago

Question 5 (20 pts) The rated current of a three-phase transmission line is 300 A. The currents flowing by the line are measured

prisoha [69]

Answer:

Check the explanation

Explanation:

Question 1.

The secondary current of 250/5 amps CT when 300 amps(rated current of transmission line ) flow in TL is

(5/250 ) X 300 = 6 amps

Question 2

The correct answer to this second question is yes, when Over current relay coil will operate and relay contacts gets close, if the pickup value( Ip) of relay is set as 6 amps in relay. ( because primary current of TL is 1.2 times of CT primary)

Question 3

Tap Block figure (Fig 1) is not available/uploaded in your question.

3 0

3 years ago