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Basile [38]
3 years ago
15

25 An 8-bit computer uses 2's complement to store negative numbers. The range of numbers that can be

Engineering
1 answer:
Inga [223]3 years ago
5 0

Answer:

-128 to +127

Explanation:

8-bits = 2⁸ = 256 possible values

Half of those will be positive and the other half will be negative.

So 128 positive number and -128 negative numbers. But we must also include

the value 0, so now the range will be from -128 to 127

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A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 3
aleksandrvk [35]

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to  250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>

<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table  " properties of air "  contained in your textbook</em>

average convection heat transfer coefficient = 25.04 W/m^2 .k

<u>ii) Determine how long this process has taken </u>

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

3 0
3 years ago
1. How many pieces of 12-1/2" long wood can be cut from a piece of
Pani-rosa [81]
B. If you do 39-3/4 divided by 12-1/2 you get 3.18
8 0
3 years ago
Select the correct answer. Sean is a recent engineering graduate who has joined a new company. Read the profiles of his colleagu
PSYCHO15rus [73]

Answer:

giving advertising to the governor with the state

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3 years ago
A scale model is 4th the size of the pump. Determine the power ratio of the pump and its scale model if the ratio of the heads i
vredina [299]

Given:

size of scale model = 4(size of pump)

power ratio of pump and scale model = 5:1

Solution:

Let the diameter of scale model and pump be d_{s} and d_{p} respectively

and head be  H_{s} and  H_{p} respectively

Now, power, P is given as a function of head(H) and dischagre(Q)

P = \rho gQH                  (1)

From eqn (1):

P \propto QH

and

QH \propto \sqrt{H}D^{2}

So,

P \propto H^{\frac{3}{2}} D^{2}

Therefore,

\frac{P_{s}}{P_{p}} = \frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}

\frac{P_{s}}{P_{p}} = \frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}

\frac{P_{s}}{P_{p}} = \frac{1^{2}\times 5^{\frac{3}{2}}}{4^{2}\times 1^{\frac{3}{2}}}

\frac{P_{s}}{P_{p}} = \frac{5\sqrt{5}}{16}

{P_{s}}:{P_{p}} = {5\sqrt{5}}:{16}

8 0
3 years ago
A 4 cm diameter sphere of copper is initially at a temperature of 95 °C. It is placed in a very large water bath at time t- 0. T
avanturin [10]

Answer:112.376 s

Explanation:

Given

T_i=95^{\circ}C

T_f=35^{\circ}C

T_\infty \left ( ambient\right )=25^{\circ}C

h=400 watts/\left ( m^{2}^{\circ}C\right )

c=0.385 J/\left ( m^2^{\circ}C\right )

\rho =9 gm/cm^{3}

Using Newton's law of cooling

\frac{T_i-T_{\infty}}{T-T_{\infty}}=e^{\frac{ht}{\rho L_{c}c}}

\frac{95-25}{35-25}=e^{\frac{400\times 3\times 10^{-4}\times t}{9\times 2\times 0.385}}

7=e^{1.7316\times 10^{-2}\times t}

Taking log both side

t=112.376sec

4 0
3 years ago
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