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katrin2010 [14]
3 years ago
15

Given a dictionary d and a list lst, remove all elements from the dictionary whose key is an element of lst. For example, given

the dictionary {1:2, 3:4, 5:6, 7:8} and the list [1, 7], the resulting dictionary would be {3:4, 5:6}. Assume every element of the list is a key in the dictionary.
Engineering
1 answer:
NikAS [45]3 years ago
6 0

Answer:

d = {1:2, 3:4, 5:6, 7:8}

list = [1, 7]

not_found = set()

for x in list:

  if x not in d.keys():

     not_found.add(x)

  else:

     del d[x]

print(d)

print(not_found)

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A force of 16,000 will cause a 1 1 bar of magnesium to stretch from 10 to 10.036 . Calculate the modulus of elasticity in . (Ent
Bad White [126]

The modulus of elasticity is 44.4GPa

E<u>xplanation:</u>

Given

original length L1=10cm

New length L2=10.036 cm

Force F=16000N

dimensions\ of\ the\ bar\ is\ 1 cm\times1cm\\Area=1cm\times1cm=1cm^2=0.0001m^2

Stress of the bar σ=Force/Area

=16000/0.0001=16\times 10^7N/m^2

Change in length ΔL=L2-L1=10.036-10=0.036 cm

Strain is obtained by dividing the change in length by original length

Strain ε=ΔL/L1

=0.036/10=0.0036

Modulus of elasticity=stress/strain

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5 0
3 years ago
The transfer function of a typical tape-drive system is given by
maw [93]

Answer:

the range of K can be said to be :  -3.59 < K< 0.35

Explanation:

The transfer function of a typical tape-drive system is given by;

KG(s) = \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]}

calculating the characteristics equation; we have:

1 + KG(s) = 0

1+   \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]} = 0

{s[s+0.5)(s+1)(s^2+0.4s+4)]} +{K(s+4)}= 0

s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ 2s+K(s+4) = 0

s^5 + 1.9 s^4+ 5.1s^3+6.2s^2+ (2+K)s+ 4K = 0

We can compute a Simulation Table for the Routh–Hurwitz stability criterion Table as  follows:

S^5             1                     5.1                          2+ K

S^4            1.9                   6.2                           4K

S^3             1.83            \dfrac{1.9 (2+K)-4K}{1.9}          0

S^2        \dfrac{11.34-1.9(X)}{1.83}       4K                         0

S          \dfrac{XY-7.32 \ K}{Y}        0                            0

\dfrac{1.9 (2+K)-4K}{1.9} = X

 

\dfrac{11.34-1.9(X)}{1.83}= Y

We need to understand that in a given stable system; all the elements in the first column is usually greater than zero

So;

11.34  - 1.9(X) > 0

11.34  - 1.9(\dfrac{3.8+1.9K-4K}{1.9}) > 0

11.34  - (3.8 - 2.1K)>0

7.54 +2.1 K > 0

2.1 K > - 7.54

K > - 7.54/2.1

K > - 3.59

Also

4K >0

K > 0/4

K > 0

Similarly;

XY - 7.32 K > 0

(\dfrac{3.8+1.9K-4K}{1.9})[11.34  - 1.9(\dfrac{3.8+1.9K-4K}{1.83}) > 7.32 \ K]

0.54(2.1K+7.54)>7.32 K

11.45 K < 4.07

K < 4.07/11.45

K < 0.35

Thus the range of K can be said to be :  -3.59 < K< 0.35

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Explanation:

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Explanation:

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