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dalvyx [7]
3 years ago
5

Assume that our computer stores decimal numbers using 16 bits - 10 bits for a sign/magnitude mantissa and 6 bits for a sign/magn

itude base-2 exponent. (The way we showed in class.) Show the internal representation of the following decimal quantities.
a. +7.5
b. -20.25
c. -1/64
Physics
1 answer:
madreJ [45]3 years ago
5 0

Explanation:

a) 7.5= 111.1×2°= 0.1111×2^3

which can also be written as

(1/2+1/4+1/8+1/16)×8

sign of mantissa:=0

Mantissa(9 bits): 111100000

sign of exponent: 0

Exponent(5 bits): 0011

the final for this is:011110000000011

b) -20.25= -10100.01×2^0= -0.1010001×2^5

sign of mantissa: 1

Mantissa(9 bits): 101000100

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1101000100000101

c)-1/64= -.000001×2^0= -0.1×2^{-5}

sign of mantissa: 1

Mantissa(9 bits): 100000000

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1100000000100101

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At a time when mining asteroids has become feasible, astronauts have connected a line between their 3740-kg space tug and a 5690
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Answer:

61.4 s

Explanation:

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d_1=\frac{1}{2}a_1t^2

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d_2=\frac{1}{2}a_2t^2

The total distance d:

d=d_1+d_2=\frac{1}{2}(a_1+a_2)t^2

Solving for time t:

t=\sqrt{\frac{2d}{a_1+a_2}}=\sqrt{\frac{2d}{F(\frac{1}{m_1}+\frac{1}{m_2})}}=\sqrt{\frac{2dm_1m_2}{F(m_1+m_2)}}

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During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s
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Answer:

part a : <em>The dry unit weight is 0.0616  </em>lb/in^3<em />

part b : <em>The void ratio is 0.77</em>

part c :  <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}

Here

  • \gamma_d is the dry unit weight which is to be calculated
  • γ is the bulk unit weight given as

                                              \gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3

  • w is the moisture content in percentage, given as 12%

Substituting values

                                              \gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3

<em>The dry unit weight is 0.0616  </em>lb/in^3<em />

Part b

Void Ratio

The void ratio is given as

                                                e=\frac{G_s \gamma_w}{\gamma_d} -1

Here

  • e is the void ratio which is to be calculated
  • \gamma_d is the dry unit weight which is calculated in part a
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72

Substituting values

                                              e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

S=\frac{G w}{e}

Here

  • e is the void ratio which is calculated in part b
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}

Here

  • \gamma_{zav} is  the zero air unit weight which is to be calculated
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Now as the volume is known, the the overall weight is given as

weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

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